5.OSCILLATIONS 12TH PHYSICS CET MCQ

Oscillation MCQ Practice

Physics | MHT-CET Previous year questions with detailed solutions

MHT-CET

Q1. Two springs of constants k₁ and k₂ have equal maximum velocities, when executing simple harmonic motion. The ratio of their amplitudes (masses are equal) will be:

(A) \(\frac{k_1}{k_2}\)

(B) \(\frac{k_2}{k_1}\)

(C) \(\left(\frac{k_1}{k_2}\right)^{1/2}\)

(D) \(\left(\frac{k_2}{k_1}\right)^{1/2}\)

Correct Answer

(D) \(\left(\frac{k_2}{k_1}\right)^{1/2}\)

Solution: The maximum velocity (\(v_{max}\)) in SHM is given by: \[v_{max} = A\omega\] Where \(\omega = \sqrt{\frac{k}{m}}\). Since the masses (\(m\)) and maximum velocities are equal for both springs: \[A_1\sqrt{\frac{k_1}{m}} = A_2\sqrt{\frac{k_2}{m}}\] Squaring both sides and canceling \(m\): \[A_1^2 k_1 = A_2^2 k_2 \implies \frac{A_1^2}{A_2^2} = \frac{k_2}{k_1}\] Taking the square root: \[\frac{A_1}{A_2} = \sqrt{\frac{k_2}{k_1}} = \left(\frac{k_2}{k_1}\right)^{1/2}\]

MHT-CET

Q2. The graph between the time period and the length of a simple pendulum is:

(A) Straight line

(B) Curve

(C) Circle

(D) Ellipse

Correct Answer

(B) Curve

Solution: The time period \(T\) of a simple pendulum is: \[T = 2\pi\sqrt{\frac{L}{g}}\] Squaring both sides gives: \[T^2 = \frac{4\pi^2}{g} L\]

• If the graph was between \(T^2\) and \(L\), it would be a straight line.
• However, the relationship between \(T\) and \(L\) is \(T \propto \sqrt{L}\). This represents a parabolic curve (specifically, \(L\) is proportional to \(T^2\)).

MHT-CET

Q3. A particle is executing S.H.M. of periodic time 'T'. The time taken by a particle in moving from mean position to half the maximum displacement is (sin 30° = 0.5):

(A) \(\frac{T}{8}\)

(B) \(\frac{T}{6}\)

(C) \(\frac{T}{4}\)

(D) \(\frac{T}{12}\)

Correct Answer

(D) \(\frac{T}{12}\)

Solution: The equation for displacement from the mean position is: \[x = A\sin(\omega t)\] Given \(x = \frac{A}{2}\) and \(\omega = \frac{2\pi}{T}\): \[\frac{A}{2} = A\sin\left(\frac{2\pi}{T} t\right) \implies \sin\left(\frac{2\pi}{T} t\right) = \frac{1}{2}\] We know \(\sin(30^\circ)\) or \(\sin(\pi/6) = 1/2\): \[\frac{2\pi}{T} t = \frac{\pi}{6}\] \[t = \frac{T}{12}\]

MHT-CET

Q4. A mass is suspended from a spring having spring constant 'K' is displaced vertically and released, it oscillates with period 'T'. The weight of the mass suspended is (g = gravitational acceleration):

(A) \(\frac{KT^2}{2\pi g}\)

(B) \(\frac{KT^2g}{4\pi^2}\)

(C) \(\frac{2\pi KT}{g}\)

(D) \(\frac{KT}{2\pi g}\)

Correct Answer

(B) \(\frac{KT^2g}{4\pi^2}\)

Solution: The time period of a mass-spring system is: \[T = 2\pi\sqrt{\frac{m}{K}}\] We need to find the weight \(W = mg\). First, solve for \(m\): \[T^2 = 4\pi^2 \frac{m}{K} \implies m = \frac{KT^2}{4\pi^2}\] Now, multiply by \(g\) to find the weight: \[W = mg = \frac{KT^2g}{4\pi^2}\]

MHT-CET

Q5. A simple pendulum is oscillating with amplitude 'A' and angular frequency 'ω'. At displacement 'x' from mean position, the ratio of kinetic energy to potential energy is:

(A) \(\frac{x^2}{A^2 - x^2}\)

(B) \(\frac{x^2 - A^2}{x^2}\)

(C) \(\frac{A^2 - x^2}{x^2}\)

(D) \(\frac{A - x}{x}\)

Correct Answer

(C) \(\frac{A^2 - x^2}{x^2}\)

Solution:

• Potential Energy (P.E.) at displacement x is: P.E. = \(\frac{1}{2}m\omega^2 x^2\)
• Kinetic Energy (K.E.) at displacement x is: K.E. = \(\frac{1}{2}m\omega^2 (A^2 - x^2)\)
• The ratio \(\frac{K.E.}{P.E.}\) is: \[\frac{\frac{1}{2}m\omega^2 (A^2 - x^2)}{\frac{1}{2}m\omega^2 x^2} = \frac{A^2 - x^2}{x^2}\]

MHT-CET 2015

Q6. A particle performs S.H.M. with amplitude 25 cm and period 3 s. The minimum time required for it to move between two points 12.5 cm on either side of the mean position is:

(A) 0.6 s

(B) 0.5 s

(C) 0.4 s

(D) 0.2 s

Correct Answer

(B) 0.5 s

Solution:

• Moving "between two points 12.5 cm on either side" means the particle moves from x = -12.5 cm to x = +12.5 cm.
• Time to go from mean position (x = 0) to x = 12.5 cm: x = A sin(ωt)
• 12.5 = 25 sin(\(\frac{2\pi}{T} t\)) ⇒ sin(\(\frac{2\pi}{3} t\)) = \(\frac{1}{2}\)
• Since sin(30°) = 1/2, then \(\frac{2\pi}{3}t = \frac{\pi}{6}\) ⇒ t = 0.25 s
• Total time for both sides = 2 × 0.25 s = 0.5 s

MHT-CET 2016

Q7. A simple pendulum of length 'l' has maximum angular displacement 'θ'. The maximum kinetic energy of the bob of mass 'm' is (g = acceleration due to gravity):

(A) mgl(1 + cos θ)

(B) mgl(1 + cos² θ)

(C) mgl(1 - cos θ)

(D) mgl(cos θ - 1)

Correct Answer

(C) mgl(1 - cos θ)

Solution:

• Maximum K.E. is equal to the Potential Energy at the maximum height (h).
• From trigonometry, the height h reached by the bob is: h = l - l cos θ = l(1 - cos θ)
• K.E._max = mgh = mgl(1 - cos θ)

MHT-CET 2016

Q8. Which of the following quantity does NOT change due to damping of oscillations?

(A) Angular frequency

(B) Time period

(C) Initial phase

(D) Amplitude

Correct Answer

(C) Initial phase

Solution:

• Damping forces (like air resistance) cause the Amplitude to decrease over time.
• Consequently, the Total Energy and Maximum Velocity also change.
• In a standard damped harmonic oscillator, the Initial Phase (which depends only on starting conditions) remains a constant parameter of the motion's description.

MHT-CET 2017

Q9. A particle is performing S.H.M. starting from extreme position. Graphical representation shows that, between displacement and acceleration, there is a phase difference of:

(A) 0 rad

(B) \(\frac{\pi}{4}\) rad

(C) \(\frac{\pi}{2}\) rad

(D) \(\pi\) rad

Correct Answer

(D) \(\pi\) rad

Solution:

• Displacement: x = A cos(ωt) (starting from extreme)
• Acceleration: a = -ω² x = -ω² A cos(ωt)
• Using the identity -cos(θ) = cos(θ + π), the acceleration can be written as: a = ω² A cos(ωt + π)
• The phase difference is π radians.

MHT-CET 2017

Q10. A solid sphere of mass 2 kg is rolling on a frictionless horizontal surface with velocity 6 m/s. It collides on the free end of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be (Force constant of the spring = 36 N/m):

(A) √14 m

(B) √2.8 m

(C) √1.4 m

(D) √0.7 m

Correct Answer

(B) √2.8 m

Solution:

• Total Kinetic Energy of a rolling sphere = Translational K.E. + Rotational K.E.
• K.E._total = \(\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{7}{10}mv^2\)
• Using values: K.E._total = \(\frac{7}{10}(2)(6)^2 = \frac{7}{10}(72) = 50.4\text{ J}\)
• Energy conservation: K.E._total = Spring P.E. = \(\frac{1}{2}kx^2\)
• 50.4 = \(\frac{1}{2}(36)x^2\) ⇒ 50.4 = 18x² ⇒ x² = 2.8 ⇒ x = √2.8 m

MHT-CET 2018

Q11. A mass is suspended from a vertical spring which is executing S.H.M. of frequency 5 Hz. The spring is unstretched at the highest point of oscillation. Maximum speed of the mass is (acceleration due to gravity g = 10 m/s²):

(A) 2π m/s

(B) π m/s

(C) \(\frac{1}{2\pi}\) m/s

(D) \(\frac{1}{\pi}\) m/s

Correct Answer

(D) \(\frac{1}{\pi}\) m/s

Solution:

• If the spring is unstretched at the highest point, the amplitude A is equal to the static extension e.
• At equilibrium, mg = ke ⇒ \(\frac{m}{k} = \frac{e}{g}\)
• Since ω = \(\sqrt{\frac{k}{m}}\), we have ω² = \(\frac{g}{e}\). Because A = e, ω² = \(\frac{g}{A}\) ⇒ A = \(\frac{g}{\omega^2}\)
• Maximum speed v_max = Aω = \((\frac{g}{\omega^2})\omega = \frac{g}{\omega}\)
• Given f = 5 Hz, ω = 2πf = 10π
• v_max = \(\frac{10}{10\pi} = \frac{1}{\pi}\) m/s

MHT-CET 2015

Q12. For a particle performing linear S.H.M., its average speed over one oscillation is (a = amplitude of S.H.M., n = frequency of oscillation):

(A) 2an

(B) 4an

(C) 6an

(D) 8an

Correct Answer

(B) 4an

Solution:

• Average speed is defined as the total distance traveled divided by the total time taken.
• In one complete oscillation, a particle starts from the mean position, goes to one extreme (a), returns to the mean (a), goes to the other extreme (a), and returns to the mean (a).
• Total distance = a + a + a + a = 4a
• Total time is the period T = 1/n
• Average speed = \(\frac{\text{Total Distance}}{\text{Total Time}} = \frac{4a}{1/n} = \mathbf{4an}\)

MHT-CET 2018

Q13. The path length of oscillation of a simple pendulum of length 1 metre is 16 cm. Its maximum velocity is (g = π² m/s²):

(A) 2π cm/s

(B) 4π cm/s

(C) 8π cm/s

(D) 16π cm/s

Correct Answer

(C) 8π cm/s

Solution:

• Path length is the total distance between the two extremes, which is 2A.
• Therefore, 2A = 16 cm ⇒ A = 8 cm
• The angular frequency ω for a pendulum is \(\sqrt{g/l}\)
• Given l = 1 m and g = π², then ω = \(\sqrt{\pi^2/1} = \pi\) rad/s
• Maximum velocity v_max = Aω = 8 cm × π rad/s = 8π cm/s

MHT-CET 2019

Q14. A weightless spring which has a force constant 'k' oscillates with frequency 'n' when a mass 'm' is suspended from it. The spring is cut into two equal halves and a mass 2m is suspended from one part of spring. The frequency of oscillation will now become:

(A) \(\frac{n}{2}\)

(B) 2n

(C) n

(D) n√2

Correct Answer

(C) n

Solution:

• The initial frequency is \(n = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\)
• When a spring is cut in half, the spring constant of each half doubles: k' = 2k
• The new mass is m' = 2m
• The new frequency n' is: \[n' = \frac{1}{2\pi}\sqrt{\frac{k'}{m'}} = \frac{1}{2\pi}\sqrt{\frac{2k}{2m}} = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \mathbf{n}\]

MHT-CET 2019

Q15. A particle performs S.H.M. starting from extreme position. The phase change between two successive passages through the mean position is:

(A) 2πᶜ

(B) πᶜ

(C) \(\frac{\pi^c}{2}\)

(D) \(\frac{\pi^c}{4}\)

Correct Answer

(B) πᶜ

Solution:

• "Successive passages through the mean position" means moving from the mean position traveling in one direction to the mean position traveling in the opposite direction.
• In terms of a circular motion projection, this corresponds to half a cycle or 180°.
• Phase change = π rad

MHT-CET 2019

Q16. A body performing a simple harmonic motion has potential energy 'P₁' at displacement 'x₁' and potential energy 'P₂' at displacement 'x₂'. The potential energy 'P' at displacement (x₁ + x₂) is:

(A) P₁ + P₂

(B) \(\sqrt{P_1^2 + P_2^2}\)

(C) \(P_1 + P_2 + 2\sqrt{P_1P_2}\)

(D) \(\sqrt{P_1P_2}\)

Correct Answer

(C) \(P_1 + P_2 + 2\sqrt{P_1P_2}\)

Solution:

• Potential Energy P = \(\frac{1}{2}kx^2\)
• So, P₁ = \(\frac{1}{2}kx_1^2\) ⇒ x₁ = \(\sqrt{2P_1/k}\) and P₂ = \(\frac{1}{2}kx_2^2\) ⇒ x₂ = \(\sqrt{2P_2/k}\)
• At displacement (x₁ + x₂): \[P = \frac{1}{2}k(x_1 + x_2)^2 = \frac{1}{2}k\left(\sqrt{\frac{2P_1}{k}} + \sqrt{\frac{2P_2}{k}}\right)^2\]
• Factoring out \(\sqrt{2/k}\) and squaring it gives 2/k: \[P = \frac{1}{2}k \cdot \frac{2}{k} (\sqrt{P_1} + \sqrt{P_2})^2 = (\sqrt{P_1} + \sqrt{P_2})^2\]
• Expanding the bracket: \(P = \mathbf{P_1 + P_2 + 2\sqrt{P_1P_2}}\)

MHT-CET 2019

Q17. A block of mass 'm' moving on a frictionless horizontal surface, collides with a spring of spring constant 'k' and compresses it through a distance 'x'. The maximum momentum of the block after collision is:

(A) \(\sqrt{mk} \cdot x\)

(B) zero

(C) \(\frac{kx^2}{2m}\)

(D) \(\frac{mx^2}{k}\)

Correct Answer

(A) \(\sqrt{mk} \cdot x\)

Solution:

• At maximum compression x, all kinetic energy of the block was converted into spring potential energy: K.E. = \(\frac{1}{2}kx^2\)
• Momentum (p) and Kinetic Energy (K.E.) are related by K.E. = \(\frac{p^2}{2m}\)
• Therefore, \(\frac{p^2}{2m} = \frac{1}{2}kx^2\)
• p² = mkx² ⇒ p = \(\mathbf{\sqrt{mk} \cdot x}\)

MHT-CET 2019

Q18. The quantity which does not vary periodically for a particle performing S.H.M. is:

(A) velocity

(B) acceleration

(C) total energy

(D) displacement

Correct Answer

(C) total energy

Solution:

• Velocity, Displacement, and Acceleration all vary sinusoidally (periodically) with time.
• Total Energy in S.H.M. (ignoring damping) is E = \(\frac{1}{2}kA^2\), which is a constant and does not change with time or position.

MHT-CET

Q19. The ratio of frequencies of oscillations of two simple pendulums is 3:4, then their lengths are in the ratio:

(A) 9:16

(B) 16:9

(C) 3:4

(D) 4:3

Correct Answer

(B) 16:9

Solution:

• The frequency of a simple pendulum is given by \(n = \frac{1}{2\pi} \sqrt{\frac{g}{L}}\)
• Since g is constant, frequency is inversely proportional to the square root of length: \(n \propto \frac{1}{\sqrt{L}}\)
• Therefore, \(\frac{n_1}{n_2} = \sqrt{\frac{L_2}{L_1}}\)
• Given \(\frac{n_1}{n_2} = \frac{3}{4}\), squaring both sides: \(\left(\frac{3}{4}\right)^2 = \frac{L_2}{L_1} \implies \frac{9}{16} = \frac{L_2}{L_1}\)
• The ratio of their lengths \(\frac{L_1}{L_2} = \mathbf{16:9}\)

MHT-CET

Q20. A pendulum is hanging in a uniformly rising lift. If lift's motion becomes uniformly decelerated, how will the period of oscillation change?

(A) The period will be shorter

(B) The period will remain same

(C) The period will be longer

(D) The period becomes zero

Correct Answer

(C) The period will be longer

Solution:

• The time period is \(T = 2\pi \sqrt{\frac{L}{g_{eff}}}\)
• When a lift is rising and decelerating (acceleration a is downwards), the effective gravity is \(g_{eff} = g - a\)
• Since the effective gravity decreases, the time period T (which is inversely proportional to \(\sqrt{g_{eff}}\)) will increase

MHT-CET

Q21. A small mass 'm' is suspended at the end of a wire. The frequency of oscillation for the S.H.M. along the vertical line is:

(A) \(\frac{1}{2\pi}\left(\frac{YA}{mL}\right)^{\frac{1}{2}}\)

(B) \(\frac{1}{2\pi}\left(\frac{mL}{YA}\right)^{\frac{1}{2}}\)

(C) \(\frac{1}{2\pi}\left(\frac{YA}{mL}\right)\)

(D) \(\frac{1}{2\pi}\sqrt{\frac{mYA}{L}}\)

Correct Answer

(A) \(\frac{1}{2\pi}\left(\frac{YA}{mL}\right)^{\frac{1}{2}}\)

Solution:

• Young's modulus is \(Y = \frac{FL}{A\Delta L}\). For a wire acting like a spring, the force is \(F = \left(\frac{YA}{L}\right) \Delta L\)
• This gives an equivalent spring constant \(k = \frac{YA}{L}\)
• The frequency of a mass-spring system is \(n = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\)
• Substituting k: \(n = \frac{1}{2\pi} \sqrt{\frac{YA}{mL}} = \mathbf{\frac{1}{2\pi}\left(\frac{YA}{mL}\right)^{\frac{1}{2}}}\)

MHT-CET

Q22. The displacement of the particle is \(x = 0.25 \sin(11t + 0.5)\) m. The period of S.H.M. is: (\(\pi = \frac{22}{7}\))

(A) \(\frac{2}{7}\) s

(B) \(\frac{3}{7}\) s

(C) \(\frac{4}{7}\) s

(D) \(\frac{5}{7}\) s

Correct Answer

(C) \(\frac{4}{7}\) s

Solution:

• Compare the given equation with the standard form \(x = A \sin(\omega t + \phi)\)
• Here, \(\omega = 11\) rad/s
• Since \(\omega = \frac{2\pi}{T}\), then \(T = \frac{2\pi}{\omega}\)
• \(T = \frac{2 \times (22/7)}{11} = \frac{44}{7 \times 11} = \mathbf{\frac{4}{7}\text{ s}}\)

MHT-CET

Q23. A body initially at rest is acted upon by a constant force (F) for time (t). The kinetic energy at time t is:

(A) \(\frac{F^2 t^2}{2m}\)

(B) \(\frac{F^2 t}{2m}\)

(C) \(\frac{F t^2}{2m}\)

(D) \(\frac{F^2 t^2}{m}\)

Correct Answer

(A) \(\frac{F^2 t^2}{2m}\)

Solution:

• From Newton's second law, \(F = ma\), so acceleration \(a = \frac{F}{m}\)
• Velocity after time t (starting from rest) is \(v = at = \frac{Ft}{m}\)
• Kinetic Energy K.E. = \(\frac{1}{2} mv^2 = \frac{1}{2} m \left(\frac{Ft}{m}\right)^2 = \frac{1}{2} m \frac{F^2 t^2}{m^2} = \mathbf{\frac{F^2 t^2}{2m}}\)

MHT-CET

Q24. Displacement \(x = 4 \sin \pi t\). Time taken to move from equilibrium to half the maximum displacement is:

(A) \(\frac{1}{6}\) s

(B) \(\frac{1}{4}\) s

(C) \(\frac{1}{3}\) s

(D) \(\frac{1}{2}\) s

Correct Answer

(A) \(\frac{1}{6}\) s

Solution:

• Maximum displacement (amplitude) A = 4 cm. Half the maximum displacement is x = 2 cm
• \(2 = 4 \sin(\pi t) \implies \sin(\pi t) = \frac{1}{2}\)
• Since \(\sin(30^\circ)\) or \(\sin(\pi/6) = 1/2\), then \(\pi t = \frac{\pi}{6}\)
• \(t = \mathbf{\frac{1}{6}\text{ s}}\)

MHT-CET

Q25. Body attached to a spring oscillates with frequency 'n' and total energy 'E'. If velocity in mean position is v, the spring constant is:

(A) \(\frac{2E\pi^2 n^2}{v^2}\)

(B) \(\frac{4E\pi^2 n^2}{v^2}\)

(C) \(\frac{8E\pi^2 n^2}{v^2}\)

(D) \(\frac{16E\pi^2 n^2}{v^2}\)

Correct Answer

(C) \(\frac{8E\pi^2 n^2}{v^2}\)

Solution:

• Total energy \(E = \frac{1}{2} k A^2\). Also, maximum velocity (at mean position) is \(v = A\omega\)
• From \(v = A\omega\), we get \(A = \frac{v}{\omega}\)
• Substitute A into the energy equation: \(E = \frac{1}{2} k \left(\frac{v}{\omega}\right)^2 = \frac{kv^2}{2\omega^2}\)
• We know \(\omega = 2\pi n\), so \(\omega^2 = 4\pi^2 n^2\)
• \(E = \frac{kv^2}{2(4\pi^2 n^2)} = \frac{kv^2}{8\pi^2 n^2}\)
• Rearranging for k: \(k = \mathbf{\frac{8E\pi^2 n^2}{v^2}}\)

MHT-CET 2021

Q26. A body of mass 'm' performs linear S.H.M. given by equation \(x = P \sin \omega t + Q \sin(\omega t + \pi/2)\). The total energy of the particle at any instant is:

(A) \(\frac{1}{2}m\omega^2(P^2 + Q^2 + PQ)\)

(B) \(\frac{1}{2}m\omega^2(P^2 + Q^2)\)

(C) \(\frac{1}{2}m\omega^2(P^2 + Q^2 - PQ)\)

(D) \(\frac{1}{2}m\omega^2(P + Q)^2\)

Correct Answer

(B) \(\frac{1}{2}m\omega^2(P^2 + Q^2)\)

Solution:

• The given equation is a composition of two S.H.M.s: \(x_1 = P \sin \omega t\) and \(x_2 = Q \sin(\omega t + \pi/2)\)
• The phase difference between them is \(\phi_2 - \phi_1 = \pi/2 - 0 = \pi/2\)
• The resultant amplitude R is given by: \(R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\phi_1 - \phi_2)}\)
• Substituting the values: \(R = \sqrt{P^2 + Q^2 + 2PQ \cos(\pi/2)}\)
• Since \(\cos(\pi/2) = 0\), \(R = \sqrt{P^2 + Q^2}\)
• Total energy in S.H.M. is T.E. = \(\frac{1}{2}m\omega^2 R^2\)
• Substituting \(R^2\): T.E. = \(\mathbf{\frac{1}{2}m\omega^2(P^2 + Q^2)}\)

MHT-CET 2021

Q27. A mass attached to a spring performs S.H.M. whose displacement is \(x = 3 \times 10^{-3} \cos(2\pi t)\) m. The time taken to obtain maximum speed for the first time is:

(A) 0.125 s

(B) 0.5 s

(C) 1 s

(D) 0.25 s

Correct Answer

(D) 0.25 s

Solution:

• The equation starts with a cosine function, meaning the particle starts from the extreme position at t = 0
• Maximum speed in S.H.M. occurs at the mean position
• The time taken to travel from the extreme position to the mean position is T/4
• From the given equation \(x = A \cos(\omega t)\), we have \(\omega = 2\pi\)
• Since \(\omega = 2\pi/T\), then \(2\pi = 2\pi/T \implies T = 1 \text{ s}\)
• Time required = \(T/4 = 1/4 = \mathbf{0.25 \text{ s}}\)

MHT-CET 2021, 2016

Q28. The bob of a simple pendulum performs S.H.M. in air with period 'T' and in water with period \(T_1\). Relation between 'T' and \(T_1\) is (density of bob = \(\frac{9}{8} \times 10^3\) kg/m³, density of water = 1000 kg/m³):

(A) \(T_1 = 2T\)

(B) \(T_1 = 3T\)

(C) \(T_1 = T/2\)

(D) \(T_1 = T/3\)

Correct Answer

(B) \(T_1 = 3T\)

Solution:

• The time period is \(T \propto 1/\sqrt{g}\). In water, effective gravity g' decreases due to upthrust: \(g' = g(1 - \rho_w / \rho_b)\)
• \(\rho_b = \frac{9}{8} \times 10^3 = 1125\) kg/m³, \(\rho_w = 1000\) kg/m³
• \(g' = g\left(1 - \frac{1000}{1125}\right) = g\left(1 - \frac{8}{9}\right) = g/9\)
• Ratio of periods: \(\frac{T_1}{T} = \sqrt{\frac{g}{g'}} = \sqrt{\frac{g}{g/9}} = \sqrt{9} = 3\)
• Therefore, \(\mathbf{T_1 = 3T}\)

MHT-CET 2022

Q29. If 'v' is velocity and 'a' is acceleration of a particle executing linear S.H.M., which one of the following statements is correct?

(A) when 'v' is maximum, 'a' is also maximum

(B) when 'a' is maximum, 'v' is zero

(C) when 'v' is maximum, 'a' is zero

(D) both (B) and (C)

Correct Answer

(D) both (B) and (C)

Solution:

• In S.H.M., at the extreme positions: displacement is maximum, velocity v = 0, and magnitude of acceleration a is maximum (a = ω²A)
• At the mean position: displacement is zero, acceleration a = 0, and velocity v is maximum
• Therefore:
  • When 'a' is maximum (at extremes), 'v' is zero → Statement (B) is correct
  • When 'v' is maximum (at mean position), 'a' is zero → Statement (C) is correct
• Thus, both (B) and (C) are correct

MHT-CET 2022

Q30. For a particle performing S.H.M. the equation \(\frac{d^2x}{dt^2} + \alpha x = 0\). Then the time period of the motion will be:

(A) \(2\pi \sqrt{\alpha}\)

(B) \(2\pi / \sqrt{\alpha}\)

(C) \(2\pi \alpha\)

(D) \(2\pi / \alpha\)

Correct Answer

(B) \(2\pi / \sqrt{\alpha}\)

Solution:

• The standard differential equation for S.H.M. is \(\frac{d^2x}{dt^2} + \omega^2 x = 0\)
• Comparing the given equation \(\frac{d^2x}{dt^2} + \alpha x = 0\), we find \(\omega^2 = \alpha\), so \(\omega = \sqrt{\alpha}\)
• The relationship between time period and angular frequency is \(T = \frac{2\pi}{\omega}\)
• Substituting \(\omega\): \(T = \mathbf{\frac{2\pi}{\sqrt{\alpha}}}\)

MHT-CET

Q31. An equation of a simple harmonic progressive wave is given by \(y = A \sin(100\pi t - 3x)\). The distance between two particles having a phase difference of \(\frac{\pi}{3}\) rad in metre is:

(A) \(\frac{\pi}{3}\)

(B) \(\frac{\pi}{9}\)

(C) \(\frac{\pi}{6}\)

(D) \(\frac{\pi}{12}\)

Correct Answer

(B) \(\frac{\pi}{9}\)

Solution:

• Compare the given wave equation with the standard form: \(y = A \sin(\omega t - kx)\)
• From the equation, the propagation constant \(k = 3\) rad/m
• The relationship between phase difference (\(\Delta \phi\)) and path difference (\(\Delta x\)) is: \(\Delta \phi = k \cdot \Delta x\)
• Given \(\Delta \phi = \pi/3\), substitute the values: \(\frac{\pi}{3} = 3 \cdot \Delta x\)
• Solving for \(\Delta x\): \(\Delta x = \frac{\pi}{9}\) m

MHT-CET

Q32. The maximum speed of a particle in S.H.M. is \(V\). The average speed is:

(A) \(\frac{V}{2}\)

(B) \(\frac{2V}{\pi}\)

(C) \(\frac{V}{\pi}\)

(D) \(\frac{2V}{\pi}\)

Correct Answer

(D) \(\frac{2V}{\pi}\)

Solution:

• Maximum speed (\(V\)) in S.H.M. is \(V = A\omega\)
• Average speed over one oscillation is defined as \(\frac{\text{Total Distance}}{\text{Total Time}}\)
• Total distance in one cycle = \(4A\); Total time \(T = \frac{2\pi}{\omega}\)
• Average speed \(= \frac{4A}{2\pi/\omega} = \frac{4A\omega}{2\pi} = \frac{2A\omega}{\pi}\)
• Since \(A\omega = V\), the average speed is \(\mathbf{\frac{2V}{\pi}}\)

MHT-CET

Q33. Match Column I (S.H.M. conditions) with Column II (Resultant amplitude R):

(A) (i)→(p); (ii)→(r); (iii)→(s); (iv)→(q)

(B) (i)→(q); (ii)→(p); (iii)→(s); (iv)→(r)

(C) (i)→(r); (ii)→(p); (iii)→(s); (iv)→(q)

(D) (i)→(s); (ii)→(p); (iii)→(r); (iv)→(q)

Correct Answer

(C) (i)→(r); (ii)→(p); (iii)→(s); (iv)→(q)

Solution:

• The resultant amplitude is \(R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\phi_1 - \phi_2)}\)
• (i) In phase (\(\Delta \phi = 0^\circ\)), \(A_1 = A_2 = A\): \(R = \sqrt{A^2 + A^2 + 2A^2(1)} = \sqrt{4A^2} = 2A\) → (r)
• (ii) In phase (\(\Delta \phi = 0^\circ\)), \(A_1 \neq A_2\): \(R = \sqrt{(A_1 + A_2)^2} = A_1 + A_2\) → (p)
• (iii) \(90^\circ\) out of phase, \(A_1 = A_2 = A\): \(R = \sqrt{A^2 + A^2 + 2A^2(0)} = \sqrt{2}A\) → (s)
• (iv) \(180^\circ\) out of phase, \(A_1 = A_2 = A\): \(R = \sqrt{A^2 + A^2 + 2A^2(-1)} = 0\) → (q)

MHT-CET

Q34. A small wooden cube is placed on a plank performing vertical S.H.M. of frequency \(n = \frac{3}{\pi}\) Hz. The maximum amplitude so that the block does not leave the plank is (take g = 10 m/s²):

(A) \(\frac{5}{16}\) m

(B) \(\frac{5}{18}\) m

(C) \(\frac{5}{14}\) m

(D) \(\frac{5}{12}\) m

Correct Answer

(B) \(\frac{5}{18}\) m

Solution:

• The block will leave the plank if the downward acceleration of the plank exceeds g
• Maximum acceleration in S.H.M. is \(a_{max} = A\omega^2\)
• To stay in contact: \(A\omega^2 \leq g\)
• \(\omega = 2\pi n = 2\pi \left(\frac{3}{\pi}\right) = 6\) rad/s
• \(A(6)^2 = 10 \implies 36A = 10\)
• \(A = \frac{10}{36} = \mathbf{\frac{5}{18} \text{ m}}\)

MHT-CET

Q35. For a particle performing S.H.M. of amplitude 'r', the potential energy is λ times its total energy. The displacement of the particle at that time is:

(A) \(\frac{r}{\sqrt{\lambda}}\)

(B) \(r\lambda\)

(C) \(r\sqrt{1-\lambda}\)

(D) \(r\sqrt{\lambda}\)

Correct Answer

(D) \(r\sqrt{\lambda}\)

Solution:

• Potential Energy (P.E.) = \(\frac{1}{2}kx^2\)
• Total Energy (T.E.) = \(\frac{1}{2}kr^2\) (where r is amplitude)
• Given P.E. = λ(T.E.): \(\frac{1}{2}kx^2 = \lambda \left(\frac{1}{2}kr^2\right)\)
• Canceling \(\frac{1}{2}k\): \(x^2 = \lambda r^2\)
• Taking the square root: \(x = \mathbf{r\sqrt{\lambda}}\)

MHT-CET

Q36. A body of mass 40 g is executing linear S.H.M. It experiences a force of 0.1 N when it is at a point 2.5 cm from mean position. The period and acceleration of the particle at that point is:

(A) \(\frac{\pi}{10}\) s, 2.5 m/s²

(B) \(\frac{\pi}{5}\) s, 2.5 m/s²

(C) \(\frac{\pi}{10}\) s, -2.5 m/s²

(D) \(\frac{\pi}{5}\) s, -2.5 m/s²

Correct Answer

(D) \(\frac{\pi}{5}\) s, -2.5 m/s²

Solution:

• Mass \(m = 40\text{ g} = 0.04\text{ kg}\)
• Force \(F = 0.1\text{ N}\) at displacement \(x = 2.5\text{ cm} = 0.025\text{ m}\)
• Acceleration (a): Using \(F = ma\), \(a = \frac{F}{m} = \frac{0.1}{0.04} = 2.5\text{ m/s}^2\) (negative sign indicates direction toward mean position)
• Time Period (T): In S.H.M., \(a = \omega^2 x\)
• \(2.5 = \omega^2 (0.025) \implies \omega^2 = 100 \implies \omega = 10\text{ rad/s}\)
• Since \(\omega = \frac{2\pi}{T}\), then \(T = \frac{2\pi}{10} = \frac{\pi}{5}\text{ s}\)

MHT-CET

Q37. A spring produces extension 'x' by applying a force 'F'. A body of mass 'm' suspended from spring oscillates vertically with period 'T'. The mass of the body is (Neglect mass of spring):

(A) \(\frac{T^2 F}{2\pi^2 x}\)

(B) \(\frac{T^2 F}{8\pi^2 x}\)

(C) \(\frac{T^2 F}{4\pi^2 x}\)

(D) \(\frac{T^2 F}{6\pi^2 x}\)

Correct Answer

(C) \(\frac{T^2 F}{4\pi^2 x}\)

Solution:

• Spring constant (k): From \(F = kx\), we get \(k = \frac{F}{x}\)
• Time Period (T): The formula is \(T = 2\pi \sqrt{\frac{m}{k}}\)
• Squaring both sides: \(T^2 = 4\pi^2 \frac{m}{k}\)
• Rearranging for m: \(m = \frac{T^2 k}{4\pi^2}\)
• Substitute \(k = \frac{F}{x}\): \(m = \mathbf{\frac{T^2 F}{4\pi^2 x}}\)

MHT-CET

Q38. A body of mass 'M' executes S.H.M. with frequency 'n'. At distance x from mean position, it has kinetic energy 'K' and potential energy 'P'. The amplitude of oscillation is:

(A) \(\frac{1}{\pi n} \sqrt{\frac{K+P}{2M}}\)

(B) \(\frac{1}{2\pi n} \sqrt{\frac{K+P}{M}}\)

(C) \(\frac{1}{\pi n} \sqrt{\frac{K-P}{2M}}\)

(D) \(\frac{1}{\pi n} \left( \frac{K+P}{2M} \right)^{1/2}\)

Correct Answer

(D) \(\frac{1}{\pi n} \left( \frac{K+P}{2M} \right)^{1/2}\)

Solution:

• Total Energy (E): \(E = K + P\)
• The formula for Total Energy is also \(E = \frac{1}{2} M \omega^2 A^2\)
• Angular frequency \(\omega = 2\pi n\), so \(\omega^2 = 4\pi^2 n^2\)
• Substitute \(\omega^2\) into the energy equation: \(K + P = \frac{1}{2} M (4\pi^2 n^2) A^2 = 2M\pi^2 n^2 A^2\)
• Solving for Amplitude (A): \(A^2 = \frac{K + P}{2M\pi^2 n^2}\)
• \(A = \sqrt{\frac{K + P}{2M\pi^2 n^2}} = \mathbf{\frac{1}{\pi n} \left( \frac{K+P}{2M} \right)^{1/2}}\)

MHT-CET

Q39. What will be the force constant of the spring system as shown in the figure (two springs k₁ in parallel, connected in series with k₂)?

(A) \(\frac{2k_1 k_2}{k_1 + 2k_2}\)

(B) \(\frac{2k_1 k_2}{2k_1 + k_2}\)

(C) \(\frac{k_1 k_2}{2k_1 + k_2}\)

(D) \(\frac{k_1 k_2}{k_1 + 2k_2}\)

Correct Answer

(B) \(\frac{2k_1 k_2}{2k_1 + k_2}\)

Solution:

• Parallel Part: The two k₁ springs are in parallel, so their effective constant is \(k_p = k_1 + k_1 = 2k_1\)
• Series Part: This combination is in series with k₂. The total effective constant k_eq is: \[\frac{1}{k_{eq}} = \frac{1}{k_p} + \frac{1}{k_2} = \frac{1}{2k_1} + \frac{1}{k_2}\]
• Find a common denominator: \(\frac{1}{k_{eq}} = \frac{k_2 + 2k_1}{2k_1 k_2}\)
• Inverting gives: \(k_{eq} = \mathbf{\frac{2k_1 k_2}{2k_1 + k_2}}\)

MHT-CET

Q40. A second's pendulum on Earth is taken to a planet whose mass and radius are twice that of Earth. The period of oscillation on the planet will be:

(A) \(2\sqrt{2}\) s

(B) \(2\) s

(C) \(\sqrt{2}\) s

(D) \(4\) s

Correct Answer

(A) \(2\sqrt{2}\) s

Solution:

• Gravity on Planet (gₚ): \(g = \frac{GM}{R^2}\). If M' = 2M and R' = 2R: \[g_p = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2} = \frac{g_e}{2}\]
• Time Period (T): \(T \propto \frac{1}{\sqrt{g}}\)
• Ratio: \(\frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_p}} = \sqrt{\frac{g_e}{g_e/2}} = \sqrt{2}\)
• Since \(T_e = 2\text{ s}\) (second's pendulum), then \(T_p = \mathbf{2\sqrt{2}\text{ s}}\)

MHT-CET 2023

Q41. A body of mass 200 gram is tied to a spring of spring constant 12.5 N/m. If the body moves in a circular path with constant angular speed 5 rad/s then the ratio of extension in the spring to its natural length will be:

(A) 3:2

(B) 1:2

(C) 2:3

(D) 1:3

Correct Answer

(C) 2:3

Solution:

• The spring force provides the centripetal force: \(kx = m(L + x)\omega^2\)
• Given: m = 0.2 kg, k = 12.5 N/m, ω = 5 rad/s
• Substitute: \(12.5x = 0.2(L + x)(5)^2\)
• \(12.5x = 0.2(L + x)(25) = 5(L + x)\)
• \(12.5x = 5L + 5x \implies 7.5x = 5L\)
• \(\frac{x}{L} = \frac{5}{7.5} = \frac{50}{75} = \frac{2}{3}\) ⇒ Ratio = 2:3

MHT-CET 2023

Q42. The amplitude of a particle executing S.H.M. is 3 cm. The displacement at which its kinetic energy will be 25% more than the potential energy is:

(A) 1 cm

(B) 2 cm

(C) 3 cm

(D) 4 cm

Correct Answer

(B) 2 cm

Solution:

• Condition: K.E. = P.E. + 25% of P.E. = 1.25 × P.E.
• \(\frac{1}{2}k(A^2 - x^2) = 1.25 \times \frac{1}{2}kx^2\)
• \(A^2 - x^2 = 1.25x^2 \implies A^2 = 2.25x^2\)
• Given A = 3 cm: \(9 = 2.25x^2 \implies x^2 = 4 \implies x = \mathbf{2\text{ cm}}\)

MHT-CET 2023

Q43. A particle is vibrating in S.H.M. with an amplitude of 4 cm. At what displacement is its energy half potential and half kinetic?

(A) 1 cm

(B) 2 cm

(C) 3 cm

(D) \(2\sqrt{2}\) cm

Correct Answer

(D) \(2\sqrt{2}\) cm

Solution:

• Condition: P.E. = \(\frac{1}{2}\) T.E.
• \(\frac{1}{2}kx^2 = \frac{1}{2} \times \frac{1}{2}kA^2\)
• \(x^2 = \frac{A^2}{2} \implies x = \frac{A}{\sqrt{2}}\)
• Given A = 4 cm: \(x = \frac{4}{\sqrt{2}} = \mathbf{2\sqrt{2}\text{ cm}}\)

MHT-CET 2023

Q44. Time period of simple pendulum on earth's surface is 'T'. Its time period becomes 'xT' when taken to a height R (equal to earth's radius). The value of 'x' will be:

(A) 1

(B) 2

(C) 3

(D) 4

Correct Answer

(B) 2

Solution:

• Gravity at height h: \(g_h = g\left(\frac{R}{R+h}\right)^2\)
• Given h = R: \(g_h = g\left(\frac{R}{2R}\right)^2 = \frac{g}{4}\)
• Time Period \(T \propto \frac{1}{\sqrt{g}}\)
• \(\frac{T'}{T} = \sqrt{\frac{g}{g_h}} = \sqrt{\frac{g}{g/4}} = \sqrt{4} = 2\)
• Therefore, \(x = \mathbf{2}\)

MHT-CET 2023

Q45. A body attached to a spring oscillates in horizontal plane with frequency 'n'. Its total energy is 'E' and the spring constant is 'K' then the velocity in the mean position is:

(A) \(\frac{1}{2\pi n}\sqrt{\frac{2E}{K}}\)

(B) \(2\pi n\sqrt{\frac{2E}{K}}\)

(C) \(2\pi n\sqrt{\frac{E}{2K}}\)

(D) \(\frac{1}{2\pi n}\sqrt{\frac{E}{2K}}\)

Correct Answer

(B) \(2\pi n\sqrt{\frac{2E}{K}}\)

Solution:

• Total Energy: \(E = \frac{1}{2}mv_{max}^2 \implies v_{max} = \sqrt{\frac{2E}{m}}\)
• We know \(\omega = 2\pi n = \sqrt{\frac{K}{m}} \implies m = \frac{K}{\omega^2} = \frac{K}{(2\pi n)^2}\)
• Substitute m: \(v_{max} = \sqrt{\frac{2E}{K/(2\pi n)^2}} = \sqrt{\frac{2E(2\pi n)^2}{K}} = \mathbf{2\pi n\sqrt{\frac{2E}{K}}}\)

MHT-CET

Q46. A vertical spring oscillates with period 2 s with mass 'm' suspended from it. When the mass 'm' is at rest, the spring is stretched through a distance of (Take \(g = \pi^2 = 10\)):

(A) 0.5 m

(B) 0.8 m

(C) 0.2 m

(D) 1 m

Correct Answer

(D) 1 m

Solution:

• Time Period: \(T = 2\pi\sqrt{\frac{m}{k}}\)
• At equilibrium: \(mg = ke \implies \frac{m}{k} = \frac{e}{g}\)
• Substitute: \(T = 2\pi\sqrt{\frac{e}{g}}\)
• Given T = 2 s, g = π²: \(2 = 2\pi\sqrt{\frac{e}{\pi^2}} \implies 1 = \pi \cdot \frac{\sqrt{e}}{\pi} = \sqrt{e}\)
• \(\sqrt{e} = 1 \implies e = \mathbf{1\text{ m}}\)

MHT-CET

Q47. A rubber ball filled with water, having a small hole is used as the bob of a simple pendulum. The time period of such a pendulum:

(A) keeps on increasing

(B) keeps on decreasing

(C) remains unchanged

(D) first increases and then decreases, finally having same value as at the beginning

Correct Answer

(D) first increases and then decreases, finally having same value as at the beginning

Solution:

• As water leaks out, the center of gravity of the bob moves downwards initially, increasing the effective length (l) of the pendulum
• Since \(T = 2\pi\sqrt{\frac{l}{g}}\), an increase in length causes the time period to increase
• Once the ball is nearly empty, the center of gravity moves back up to the geometric center, decreasing the length and the time period back to its original value

MHT-CET

Q48. The maximum velocity of a particle performing S.H.M. is 'V'. If the periodic time is made (1/3)rd and the amplitude is doubled, then the new maximum velocity of the particle will be:

(A) 2V

(B) 3V

(C) 4V

(D) 6V

Correct Answer

(D) 6V

Solution:

• Initial Velocity: \(V = A\omega = A\left(\frac{2\pi}{T}\right)\)
• New conditions: \(A' = 2A\) and \(T' = \frac{T}{3}\)
• New Velocity: \(V' = A'\omega' = (2A)\left(\frac{2\pi}{T/3}\right) = (2A)\left(\frac{6\pi}{T}\right)\)
• \(V' = 6 \times \left[A\left(\frac{2\pi}{T}\right)\right] = \mathbf{6V}\)

MHT-CET

Q49. Two S.H.Ms. are represented by equations \(y_1 = 0.1 \sin(100\pi t + \pi/3)\) and \(y_2 = 0.1 \cos(100\pi t)\). The phase difference between the speeds of the two particles is:

(A) \(\frac{\pi}{3}\)

(B) \(-\frac{\pi}{6}\)

(C) \(\frac{\pi}{6}\)

(D) \(-\frac{\pi}{3}\)

Correct Answer

(B) \(-\frac{\pi}{6}\)

Solution:

• Speed equations (derivatives): \(v_1 = 0.1(100\pi) \cos(100\pi t + \pi/3)\) \(v_2 = -0.1(100\pi) \sin(100\pi t)\)
• Convert both to cosine: \(v_1 \propto \cos(100\pi t + \pi/3)\) \(v_2 \propto -\sin(100\pi t) = \cos(100\pi t + \pi/2)\)
• Phase difference: \(\Delta \phi = \frac{\pi}{3} - \frac{\pi}{2} = \mathbf{-\frac{\pi}{6}}\)

MHT-CET

Q50. A particle is vibrating in S.H.M. with an amplitude of 4 cm. At what displacement from the equilibrium position is its energy half potential and half kinetic?

(A) 1 cm

(B) 2 cm

(C) 3 cm

(D) \(2\sqrt{2}\) cm

Correct Answer

(D) \(2\sqrt{2}\) cm

Solution:

• Condition: P.E. = K.E. ⇒ P.E. = \(\frac{1}{2}\) T.E.
• \(\frac{1}{2}kx^2 = \frac{1}{2} \times \frac{1}{2}kA^2\)
• \(x^2 = \frac{A^2}{2} \implies x = \frac{A}{\sqrt{2}}\)
• Given A = 4 cm: \(x = \frac{4}{\sqrt{2}} = \mathbf{2\sqrt{2}\text{ cm}}\)

MHT-CET

Q51. A second's pendulum is placed in a space laboratory orbiting around the earth at a height 3R from the earth's surface. The time period of the pendulum will be (R = radius of earth):

(A) 2 s

(B) 4 s

(C) 6 s

(D) infinite

Correct Answer

(D) infinite

Solution:

• In an orbiting satellite or space laboratory, every object is in a state of free fall towards the center of the earth
• Because the laboratory is in orbit, the effective gravitational acceleration (g_eff) inside the satellite is zero
• The formula for a simple pendulum is \(T = 2\pi\sqrt{\frac{L}{g_{eff}}}\)
• Since g_eff = 0, the expression becomes undefined - physically representing an infinite time period
• The pendulum will not oscillate at all

MHT-CET

Q52. A particle performing S.H.M. starts from equilibrium position and its time period is 12 second. After 2 seconds its velocity is π m/s. Amplitude of the oscillation is:

(A) 6 m

(B) 12 m

(C) 18 m

(D) 24 m

Correct Answer

(B) 12 m

Solution:

• Given: T = 12 s, t = 2 s, v = π m/s
• Angular Frequency: \(\omega = \frac{2\pi}{T} = \frac{2\pi}{12} = \frac{\pi}{6}\) rad/s
• Starting from equilibrium, \(x = A \sin(\omega t)\), so \(v = A\omega \cos(\omega t)\)
• \(\pi = A\left(\frac{\pi}{6}\right) \cos\left(\frac{\pi}{6} \times 2\right) = \frac{A\pi}{6} \cos\left(\frac{\pi}{3}\right)\)
• \(\cos(\pi/3) = 0.5\), so \(1 = \frac{A}{6} \times 0.5 = \frac{A}{12}\)
• \(A = \mathbf{12\text{ m}}\)

MHT-CET

Q53. A particle performs linear S.H.M. At a particular instant, velocity is u and acceleration is a₁. At another instant, velocity is v and acceleration is a₂. The distance between the two positions is:

(A) \(\frac{u^2 - v^2}{a_1 - a_2}\)

(B) \(\frac{v^2 - u^2}{a_1 - a_2}\)

(C) \(\frac{u^2 - v^2}{a_2 - a_1}\)

(D) \(\frac{u^2 - v^2}{a_1 + a_2}\)

Correct Answer

(D) \(\frac{u^2 - v^2}{a_1 + a_2}\)

Solution:

• In S.H.M., \(v^2 = \omega^2(A^2 - x^2)\) and \(a = -\omega^2 x\)
• Position 1: \(u^2 = \omega^2(A^2 - x_1^2)\) and \(x_1 = -\frac{a_1}{\omega^2}\)
• Position 2: \(v^2 = \omega^2(A^2 - x_2^2)\) and \(x_2 = -\frac{a_2}{\omega^2}\)
• \(u^2 - v^2 = \omega^2(x_2^2 - x_1^2) = \omega^2(x_2 - x_1)(x_2 + x_1)\)
• Substitute \(x_1 + x_2 = -\frac{(a_1 + a_2)}{\omega^2}\):
• \(u^2 - v^2 = \omega^2(x_2 - x_1)\left(-\frac{a_1 + a_2}{\omega^2}\right) = -(x_2 - x_1)(a_1 + a_2)\)
• Distance \((x_2 - x_1) = \mathbf{\frac{u^2 - v^2}{a_1 + a_2}}\)

MHT-CET

Q54. A simple pendulum of length 'L' has mass 'M' and oscillates with amplitude 'A'. At extreme position, its potential energy is:

(A) \(\frac{MgL^2}{2A}\)

(B) \(\frac{MgA^2}{2L}\)

(C) \(\frac{MgL}{2A^2}\)

(D) \(\frac{MgA}{2L^2}\)

Correct Answer

(B) \(\frac{MgA^2}{2L}\)

Solution:

• Restoring force constant for a simple pendulum: \(k = \frac{Mg}{L}\)
• Potential Energy at extreme position (x = A): \(P.E. = \frac{1}{2}kx^2\)
• \(P.E. = \frac{1}{2}\left(\frac{Mg}{L}\right)A^2 = \mathbf{\frac{MgA^2}{2L}}\)

MHT-CET

Q55. The displacement of a particle in S.H.M. is \(x = A \cos(\omega t + \pi/6)\). Its speed will be maximum at time:

(A) \(\frac{\pi}{6\omega}\) s

(B) \(\frac{\pi}{2\omega}\) s

(C) \(\frac{\pi}{3\omega}\) s

(D) \(\frac{\pi}{\omega}\) s

Correct Answer

(C) \(\frac{\pi}{3\omega}\) s

Solution:

• Speed is maximum when the particle is at the mean position (x = 0)
• \(0 = A \cos\left(\omega t + \frac{\pi}{6}\right)\)
• For cosine to be zero, the angle must be \(\frac{\pi}{2}\):
• \(\omega t + \frac{\pi}{6} = \frac{\pi}{2}\)
• \(\omega t = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}\)
• \(t = \mathbf{\frac{\pi}{3\omega}\text{ s}}\)

MHT-CET

Q56. A musical instrument 'P' produces sound waves of frequency 'n' and amplitude 'A'. Another musical instrument 'Q' produces sound waves of frequency n/4. The waves produced by 'P' and 'Q' have equal energies. If the amplitude of waves produced by 'P' is Aᴘ, the amplitude of waves produced by 'Q' will be:

(A) 2Aᴘ

(B) 4Aᴘ

(C) 8Aᴘ

(D) 16Aᴘ

Correct Answer

(B) 4Aᴘ

Solution:

• The energy (E) of a wave is proportional to the square of its frequency and the square of its amplitude: \(E \propto n^2 A^2\)
• Given: Eᴘ = Eǫ, so \(n_P^2 A_P^2 = n_Q^2 A_Q^2\)
• Frequency of Q is n/4: \(n^2 A_P^2 = \left(\frac{n}{4}\right)^2 A_Q^2 = \frac{n^2}{16} A_Q^2\)
• \(A_Q^2 = 16 A_P^2 \implies A_Q = \mathbf{4A_P}\)

MHT-CET

Q57. A mass attached to a horizontal spring executes S.H.M. with amplitude A₁. When the mass passes through mean position, then a mass 'm' is placed over it and both of them move together with amplitude A₂. The ratio A₂/A₁ is:

(A) \(\left(\frac{M}{M + m}\right)\)

(B) \(\left(\frac{M + m}{M}\right)^{1/2}\)

(C) \(\left(\frac{M}{M + m}\right)^{1/2}\)

(D) \(\left(\frac{M + m}{M}\right)\)

Correct Answer

(C) \(\left(\frac{M}{M + m}\right)^{1/2}\)

Solution:

• At mean position, velocity is maximum: \(v_{max} = A\omega\)
• Momentum is conserved (no external horizontal forces): \(M v_1 = (M + m) v_2\)
• Substitute velocity: \(M(A_1 \omega_1) = (M + m)(A_2 \omega_2)\)
• Angular frequencies: \(\omega_1 = \sqrt{K/M}\), \(\omega_2 = \sqrt{K/(M+m)}\)
• \(M A_1 \sqrt{\frac{K}{M}} = (M + m) A_2 \sqrt{\frac{K}{M + m}}\)
• \(A_1 \sqrt{MK} = A_2 \sqrt{(M + m)K}\)
• \(\frac{A_2}{A_1} = \sqrt{\frac{M}{M + m}} = \mathbf{\left(\frac{M}{M + m}\right)^{1/2}}\)

MHT-CET

Q58. The potential energy of a long spring when it is stretched by 3 cm is 'U'. If the spring is stretched by 9 cm, potential energy stored in it will be:

(A) 3U

(B) 6U

(C) 18U

(D) 9U

Correct Answer

(D) 9U

Solution:

• Potential Energy: \(P.E. = \frac{1}{2}kx^2\), so \(U \propto x^2\)
• Ratio: \(\frac{U_2}{U_1} = \left(\frac{x_2}{x_1}\right)^2\)
• Given x₁ = 3 cm, x₂ = 9 cm: \(\frac{U_2}{U} = \left(\frac{9}{3}\right)^2 = (3)^2 = 9\)
• \(U_2 = \mathbf{9U}\)

MHT-CET

Q59. Four massless springs each of spring constant '2K' are attached to a mass 'M' as shown. If the mass 'M' is displaced in the horizontal direction, the frequency of oscillation is:

(A) \(\frac{1}{2\pi}\sqrt{\frac{4K}{M}}\)

(B) \(\frac{1}{2\pi}\sqrt{\frac{6K}{M}}\)

(C) \(\frac{1}{2\pi}\sqrt{\frac{5K}{M}}\)

(D) \(\frac{1}{2\pi}\sqrt{\frac{8K}{M}}\)

Correct Answer

(D) \(\frac{1}{2\pi}\sqrt{\frac{8K}{M}}\)

Solution:

• When the mass is displaced horizontally, all four springs exert restoring forces in the same direction
• This means they are effectively in parallel
• Equivalent spring constant: \(K_{eq} = 2K + 2K + 2K + 2K = 8K\)
• Frequency: \(n = \frac{1}{2\pi}\sqrt{\frac{K_{eq}}{M}} = \mathbf{\frac{1}{2\pi}\sqrt{\frac{8K}{M}}}\)

MHT-CET

Q60. A small sphere oscillates in a watch glass whose radius of curvature is 1.6 m. The period of oscillation is (g = 10 m/s²):

(A) 0.4π s

(B) 0.6π s

(C) 0.2π s

(D) 0.8π s

Correct Answer

(D) 0.8π s

Solution:

• A sphere rolling in a concave surface follows the formula: \(T = 2\pi\sqrt{\frac{R - r}{g}}\)
• For a small sphere, r is negligible, so \(T \approx 2\pi\sqrt{\frac{R}{g}}\)
• Given: R = 1.6 m, g = 10 m/s²
• \(T = 2\pi\sqrt{\frac{1.6}{10}} = 2\pi\sqrt{0.16} = 2\pi(0.4) = \mathbf{0.8\pi\text{ s}}\)

MHT-CET

Q61. A block of mass 'M' is attached with spring of natural length 'L' and force constant 'K'. Now system is rotated with constant angular velocity 'ω' on a horizontal plane. The tension in the spring is:

(A) \(\frac{M\omega^2 L K}{K + M\omega^2}\)

(B) \(\frac{M\omega^2 L}{K - M\omega^2}\)

(C) \(\frac{K L}{K - M\omega^2}\)

(D) \(\frac{M\omega^2 L K}{K - M\omega^2}\)

Correct Answer

(D) \(\frac{M\omega^2 L K}{K - M\omega^2}\)

Solution:

• When the system rotates, the spring stretches by amount x. New length = L + x
• Spring force provides centripetal force: \(T = kx = M(L + x)\omega^2\)
• \(kx = ML\omega^2 + Mx\omega^2 \implies x(k - M\omega^2) = ML\omega^2\)
• \(x = \frac{ML\omega^2}{k - M\omega^2}\)
• Tension \(T = kx = \mathbf{\frac{M\omega^2 L K}{K - M\omega^2}}\)

MHT-CET

Q62. A particle is performing S.H.M. If at any moment the displacement of the particle from mean position is half of the amplitude then the ratio of its potential energy to kinetic energy is:

(A) 1:1

(B) 2:1

(C) 1:3

(D) 3:1

Correct Answer

(C) 1:3

Solution:

• Given: Displacement \(x = \frac{A}{2}\)
• P.E. = \(\frac{1}{2}kx^2 = \frac{1}{2}k\left(\frac{A}{2}\right)^2 = \frac{1}{8}kA^2\)
• K.E. = \(\frac{1}{2}k(A^2 - x^2) = \frac{1}{2}k\left(A^2 - \frac{A^2}{4}\right) = \frac{3}{8}kA^2\)
• Ratio \(\frac{P.E.}{K.E.} = \frac{1/8}{3/8} = \mathbf{1:3}\)

MHT-CET

Q63. A simple harmonic progressive wave is represented by \(y = 5 \times 10^{-4} \sin(600\pi t - \frac{\pi}{32.5}x)\). The speed of the wave and phase difference between two particles separated by 13 cm are:

(A) 195 m/s, \(\frac{2\pi}{5}\) rad

(B) 195 m/s, \(\frac{\pi}{5}\) rad

(C) 200 m/s, \(\frac{2\pi}{5}\) rad

(D) 200 m/s, \(\frac{\pi}{5}\) rad

Correct Answer

(A) 195 m/s, \(\frac{2\pi}{5}\) rad

Solution:

• Compare with \(y = A\sin(\omega t - kx)\): \(\omega = 600\pi\), \(k = \frac{\pi}{32.5}\)
• Wave speed: \(v = \frac{\omega}{k} = \frac{600\pi}{\pi/32.5} = 600 \times 32.5 = 19500\) cm/s = 195 m/s
• Phase difference: \(\Delta\phi = k \cdot \Delta x = \left(\frac{\pi}{32.5}\right) \times 13 = \frac{13\pi}{32.5} = \frac{\pi}{2.5} = \mathbf{\frac{2\pi}{5}\text{ rad}}\)

MHT-CET

Q64. A spherical marble of radius 'r' is made to oscillate in a bowl of radius 'R'. What is its period of oscillation if it is performing simple harmonic motion?

(A) \(2\pi\sqrt{\frac{R}{g}}\)

(B) \(2\pi\sqrt{\frac{R-r}{g}}\)

(C) \(2\pi\sqrt{\frac{R+r}{g}}\)

(D) \(2\pi\sqrt{\frac{r}{g}}\)

Correct Answer

(B) \(2\pi\sqrt{\frac{R-r}{g}}\)

Solution:

• For a body rolling or sliding in a hemispherical bowl, the effective radius is the distance from the center of the bowl to the center of the oscillating body
• The center of the marble is at a distance \((R - r)\) from the center of the bowl
• Time period: \(T = 2\pi\sqrt{\frac{R - r}{g}}\)

MHT-CET

Q65. Two simple pendulums A and B of lengths 1.44 m and 1.21 m start oscillating at the same time. How many oscillations pendulum B will complete when it will be out of phase with pendulum A?

(A) 5.0

(B) 5.5

(C) 6.0

(D) 6.5

Correct Answer

(C) 6.0

Solution:

• "Out of phase" means one pendulum has completed exactly half an oscillation more than the other: \(n_B = n_A + 0.5\)
• \(T \propto \sqrt{L}\), so \(T_A \propto \sqrt{1.44} = 1.2\), \(T_B \propto \sqrt{1.21} = 1.1\)
• Time taken is same: \(n_A T_A = n_B T_B\)
• \((n_B - 0.5) \times 1.2 = n_B \times 1.1\)
• \(1.2n_B - 0.6 = 1.1n_B \implies 0.1n_B = 0.6 \implies n_B = \mathbf{6.0}\)

MHT-CET

Q66. The time period of a simple pendulum is T when the length is L. When the length is increased by 10 cm, its period is T₁. When the length is decreased by 10 cm, its period is T₂. The relation between T, T₁ and T₂ is:

(A) \(T^2 = T_1^2 + T_2^2\)

(B) \(T_1^2 = T^2 + T_2^2\)

(C) \(2T^2 = T_1^2 + T_2^2\)

(D) \(T^2 = 2(T_1^2 + T_2^2)\)

Correct Answer

(C) \(2T^2 = T_1^2 + T_2^2\)

Solution:

• \(T = 2\pi\sqrt{\frac{L}{g}} \implies T^2 \propto L\)
• \(T^2 = kL\) where \(k = \frac{4\pi^2}{g}\)
• \(T_1^2 = k(L + 10) = kL + 10k\)
• \(T_2^2 = k(L - 10) = kL - 10k\)
• Adding: \(T_1^2 + T_2^2 = 2kL = 2T^2\)

MHT-CET

Q67. Three springs each of force constant 'K' are connected at equal angles with respect to each other to a common mass 'M'. The other end of the springs is rigidly fixed. If the mass is pulled towards any one of the springs, then the period of oscillation will be:

(A) \(2\pi\sqrt{\frac{M}{K}}\)

(B) \(2\pi\sqrt{\frac{2M}{3K}}\)

(C) \(2\pi\sqrt{\frac{3M}{2K}}\)

(D) \(2\pi\sqrt{\frac{M}{3K}}\)

Correct Answer

(B) \(2\pi\sqrt{\frac{2M}{3K}}\)

Solution:

• When three springs are arranged symmetrically at 120°, pulling the mass towards one spring compresses it while the other two are stretched at an angle
• The effective spring constant is \(k_{eq} = \frac{3}{2}K\)
• Time period: \(T = 2\pi\sqrt{\frac{M}{k_{eq}}} = 2\pi\sqrt{\frac{M}{1.5K}} = \mathbf{2\pi\sqrt{\frac{2M}{3K}}}\)

MHT-CET

Q68. A particle is performing simple harmonic motion and if the oscillations are damped oscillations then the angular frequency is given by:

(A) \(\sqrt{\frac{k}{m} - \left(\frac{b}{m}\right)^2}\)

(B) \(\sqrt{\frac{k}{m} - \frac{b}{2m}}\)

(C) \(\sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^2}\)

(D) \(\sqrt{\frac{k}{m} + \left(\frac{b}{2m}\right)^2}\)

Correct Answer

(C) \(\sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^2}\)

Solution:

• In damped oscillations, the restoring force is opposed by a dissipative force proportional to velocity
• The differential equation is \(m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx = 0\)
• The resulting angular frequency is lower than the natural frequency
• \(\omega' = \sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^2}\)

MHT-CET

Q69. Three masses 500 g, 300 g and 100 g are suspended at the end of a spring and are in equilibrium. When the 500 g mass is removed, the system oscillates with T = 3 s. When the 300 g mass is also removed, it will oscillate with a period of:

(A) 2.0 s

(B) 1.5 s

(C) 1.0 s

(D) 0.5 s

Correct Answer

(B) 1.5 s

Solution:

• Time period \(T \propto \sqrt{m}\)
• Case 1 (500 g removed): Remaining mass = 300 + 100 = 400 g, T₁ = 3 s
• Case 2 (300 g also removed): Remaining mass = 100 g, T₂ = ?
• \(\frac{T_2}{T_1} = \sqrt{\frac{100}{400}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\)
• \(T_2 = \frac{3}{2} = \mathbf{1.5\text{ s}}\)

MHT-CET

Q70. The string of a pendulum of length 'L' is displaced through 90° from the vertical and released. The maximum tension the string must withstand as the pendulum passes through the mean position is:

(A) 2mg

(B) 3mg

(C) mg

(D) 4mg

Correct Answer

(B) 3mg

Solution:

• At 90° position, height h = L, P.E. = mgL
• At mean position, all P.E. converts to K.E.: \(\frac{1}{2}mv^2 = mgL \implies v^2 = 2gL\)
• Tension at mean position must support weight and provide centripetal force:
• \(T = mg + \frac{mv^2}{L} = mg + \frac{m(2gL)}{L} = mg + 2mg = \mathbf{3mg}\)

MHT-CET

Q71. The displacement of a wave is given by \(y = 0.002 \sin(100t + x)\) where x and y are in metre and t is in second. This represents a wave:

(A) travelling with a velocity of 100 m/s in the positive x-direction

(B) travelling with a velocity of 100 m/s in the negative x-direction

(C) travelling with a velocity of 50 m/s in the positive x-direction

(D) travelling with a velocity of 50 m/s in the negative x-direction

Correct Answer

(B) travelling with a velocity of 100 m/s in the negative x-direction

Solution:

• Compare with standard wave equation \(y = A \sin(\omega t + kx)\)
• Since the sign between ωt and kx is positive (+), the wave is travelling in the negative x-direction
• From the equation: ω = 100 rad/s, k = 1 rad/m
• Velocity: \(v = \frac{\omega}{k} = \frac{100}{1} = \mathbf{100\text{ m/s}}\)
• Therefore, the wave travels at 100 m/s in the negative x-direction

MHT-CET

Q72. Frequency of a particle performing S.H.M. is 10 Hz. The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is unstretched. Maximum speed of the particle is (g = 10 m/s²):

(A) \(\frac{1}{\pi}\) m/s

(B) \(\frac{1}{2\pi}\) m/s

(C) \(\frac{2}{\pi}\) m/s

(D) \(\frac{1}{4\pi}\) m/s

Correct Answer

(B) \(\frac{1}{2\pi}\) m/s

Solution:

• If spring is unstretched at highest point, amplitude A = static extension e
• At equilibrium: mg = ke ⇒ \(\frac{k}{m} = \frac{g}{e}\)
• Angular frequency: \(\omega^2 = \frac{k}{m} = \frac{g}{A}\)
• Maximum speed: \(v_{max} = A\omega = \left(\frac{g}{\omega^2}\right)\omega = \frac{g}{\omega}\)
• Given f = 10 Hz, ω = 2πf = 20π
• \(v_{max} = \frac{10}{20\pi} = \mathbf{\frac{1}{2\pi}\text{ m/s}}\)

MHT-CET

Q73. In S.H.M. the displacement of a particle at an instant is \(y = A \cos 30^\circ\), where A = 40 cm and kinetic energy is 200 J. If force constant is \(1 \times 10^x\) N/m, then x will be:

(A) 4

(B) 2

(C) 3

(D) 5

Correct Answer

(A) 4

Solution:

• K.E. = \(\frac{1}{2}k(A^2 - y^2)\)
• Given: y = A cos30° = A × \(\frac{\sqrt{3}}{2}\), so \(y^2 = A^2 \times \frac{3}{4}\)
• 200 = \(\frac{1}{2}k\left(A^2 - \frac{3}{4}A^2\right) = \frac{1}{2}k \times \frac{A^2}{4} = \frac{kA^2}{8}\)
• A = 40 cm = 0.4 m, so 200 = \(\frac{k(0.4)^2}{8} = \frac{k \times 0.16}{8} = 0.02k\)
• k = \(\frac{200}{0.02} = 10000 = 1 \times 10^4\) N/m
• Therefore, x = 4

MHT-CET

Q74. Let l₁ be the length of simple pendulum. Its length changes to l₂ to increase the periodic time by 20%. The ratio l₂/l₁ = :

(A) 1.20

(B) 1.44

(C) 1.69

(D) 1.96

Correct Answer

(B) 1.44

Solution:

• Time period relation: \(T \propto \sqrt{l} \implies \frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}}\)
• Given: T₂ = T₁ + 20% of T₁ = 1.2 T₁
• \(\frac{1.2 T_1}{T_1} = \sqrt{\frac{l_2}{l_1}} \implies 1.2 = \sqrt{\frac{l_2}{l_1}}\)
• Squaring both sides: \(\frac{l_2}{l_1} = (1.2)^2 = \mathbf{1.44}\)

MHT-CET

Q75. A piece of wood (length a, breadth b, height c) has relative density 'd'. It is floating in water such that side 'a' is vertical. If pushed down a little and released, the time period of S.H.M. is:

(A) \(2\pi\sqrt{\frac{bc}{g}}\)

(B) \(2\pi\sqrt{\frac{abd}{g}}\)

(C) \(2\pi\sqrt{\frac{ac}{g}}\)

(D) \(2\pi\sqrt{\frac{ad}{g}}\)

Correct Answer

(D) \(2\pi\sqrt{\frac{ad}{g}}\)

Solution:

• When pushed down by x, extra upthrust: \(F = -(bc \cdot \rho_w \cdot g)x\)
• Spring constant: \(k = bc \rho_w g\)
• Mass of wood: \(m = abc \cdot d\rho_w\)
• Time period: \(T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{abcd\rho_w}{bc\rho_w g}} = \mathbf{2\pi\sqrt{\frac{ad}{g}}}\)

MHT-CET

Q76. A string is under tension 180 N and mass per unit length \(2 \times 10^{-3}\) kg/m. It produces consecutive resonant frequencies 375 Hz and 450 Hz. The mass of the string is:

(A) 2 gram

(B) 8 gram

(C) 6 gram

(D) 4 gram

Correct Answer

(D) 4 gram

Solution:

• Fundamental frequency: \(n_0 = 450 - 375 = 75\) Hz
• Wave velocity: \(v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{180}{2 \times 10^{-3}}} = \sqrt{90,000} = 300\) m/s
• Length: \(n_0 = \frac{v}{2L} \implies 75 = \frac{300}{2L} \implies 2L = \frac{300}{75} = 4 \implies L = 2\) m
• Mass of string: \(M = \mu \times L = (2 \times 10^{-3}) \times 2 = 4 \times 10^{-3}\) kg = 4 gram