MHT-CET PYQs 2024 - Physics: Oscillations

OSCILLATIONS

1. The motion of a particle is described by the equation a = -bx where ‘a’ is the acceleration, x is the displacement from the equilibrium position and b is a constant. The periodic time will be

[MHT-CET 2024, May 15, Shift 2]

  • (1) \( \frac{2\pi}{b} \)
  • (2) \( \frac{2\pi}{\sqrt{b}} \)
  • (3) \( {2\pi\sqrt{b}} \)
  • (4) \( 2{\sqrt{\frac{\pi}{b}}} \)

 

2. When a particle in linear S.H.M. completes two oscillations, its phase increases by

[MHT-CET 2024, May 3, Shift 2]

(1) π \( rad \)
(2) 2π \( rad \)
(3) 3π \( rad \)
(4) 4π \( rad \)

 

3. The maximum velocity and maximum acceleration of a particle performing a linear S.H.M. is ‘α’ and ‘β’ respectively. Then the path length of the particle is

[MHT-CET 2024, May 4, Shift 2]

(1)\( \frac{α^2}{β} \)
(2)\( \frac{β}{2α^2} \)
(3)\( \frac{2α^2}{β}\)
(4)\( \frac{2β}{α^2}\)

 

4. A particle is performing S.H.M. with maximum velocity ‘V’. If the amplitude is doubled and periodic time is made \( \left(\frac{1}{3}\right)^{rd} \) then the maximum velocity is

[MHT-CET 2024, May 10, Shift 1]

(1) \(\frac{V}{2}\)
(2) \(\frac{V}{3}\)
(3) \(\frac{3V}{2}\)
(4)\(\frac{2V}{3}\)

 

5. For a particle executing S.H.M. having amplitude A, the speed of the particle is \( \left(\frac{1}{3}\right)^{rd} \) of its maximum speed when the displacement from the mean position is

[MHT-CET 2024, May 15, Shift 2]

(1) \(\frac{3A}{\sqrt 2}\)
(2) \(\frac{2A}{3}\)
(3) \(\frac{2 \sqrt 2}{3}\) A
(4) \(\frac{\sqrt 2}{3}\) A

 

6. A particle is performing S.H.M. about its mean position with an amplitude ‘a’ and periodic time ‘T’. The speed of the particle when its displacement from mean position is \(\frac{a}{3}\) will be

[MHT-CET 2024, May 11, Shift 2]

(1) \(\frac{2πa}{T}\)
(2) \(\frac{4\sqrt2 πa}{3T}\)
(3) \(\frac{4π^2a}{3T}\)
(4) \(\frac{\sqrt3 π^2a}{2T}\)

 

7. A particle performing S.H.M. starts from equilibrium position and its time period is 12 second. After 2 seconds its velocity is πm/s. Amplitude of the oscillation is [sin 30º = cos 60º = 0.5, sin 60º = cos 30º = \( \sqrt{3}/2 \) ]

[MHT-CET 2024, May 11, Shift 1]

(1) 6 m
(2) 12 m
(3) \( 12\sqrt{3} \)m
(4) \( 6\sqrt{3} \)m

 

8. The velocity of particle executing S.H.M. varies with displacement (x) as \( 4V^2 \) = 50 – \( x^2\). The time period of oscillation is \(\frac{x}{7}\) second. The value of ‘x’ is (Take π = \(\frac{22}{7}\) )

[MHT-CET 2024, May 4, Shift 1]

(1) 22
(2) 44
(3) 66
(4) 88

 

9. A particle is performing S.H.M. with an amplitude 4 cm. At the mean position the velocity of the particle is 12 cm/s. When the speed of the particle becomes 6 cm/s, the distance of the particle from mean position is

[MHT-CET 2024, May 9, Shift 1]

(1) \( \sqrt{3} \) cm
(2) \( \sqrt{6} \) cm
(3) 2\( \sqrt{3} \) cm
(4) 2\( \sqrt{6} \) cm

 

10. A particle executing S.H.M. has velocities ‘\( V_1 \)’ and ‘\( V_2 \)’ at distances ‘\( x_1 \)’ and ‘\( x_2 \)’ respectively, from the mean position. Its frequency is

[MHT-CET 2024, May 16, Shift 1]

(1) \( \frac{1}{2\pi} \sqrt{\frac{V_1^2 - V_2^2}{x_1^2 - x_2^2}} \)
(2) \( {2\pi} \sqrt{\frac{x_1^2 - x_2^2}{V_1^2 - V_2^2}} \)
(3) \( \frac{1}{2\pi} \sqrt{\frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}} \)
(4) \( {2\pi} \sqrt{\frac{x_2^2 - x_1^2}{V_2^2 - V_1^2}} \)

 

11. A particle is executing a linear simple harmonic motion. Let ‘\( V_1 \) ’ and ‘\( V_2 \) ’ are its speed at distance ‘\( x_1 \) ’ and ‘\( x_2 \) ’ from the equilibrium position. The amplitude of oscillation is

[MHT-CET 2024, May 9, Shift 2]

(1) \( \sqrt{\frac{V_1^2 x_2^2 - V_2^2 x_2^2}{V_1^2 - V_2^2}} \)
(2) \( \sqrt{\frac{V_1^2 - V_2^2}{V_1^2 x_2^2 - V_2^2 x_1^2}} \)
(3) \( \sqrt{\frac{V_1^2 x_2^2 - V_2^2 x_1^2}{V_1^2 - V_2^2}} \)
(4) \( \sqrt{\frac{V_1^2 x_1^2 - V_2^2 x_2^2}{V_1^2 - V_2^2}} \)

 

12. A particle performs linear S.H.M. When the displacement of the particle from mean position is 3 cm and 4 cm, corresponding velocities are 8 cm/s and 6 cm/s respectively. Its periodic time is

[MHT-CET 2024, May 10, Shift 2]

(1) 2π s
(2) π s
(3) 3π s
(4) 4π s

 

13. A particle performs linear S.H.M. At a particular instant, the velocity of the particle is ‘u’ and the acceleration is ‘\( a_1 \)’, while at another instant the velocity is ‘V’ and the acceleration is ‘\( a_2 \)‘. (\( 0 < a_1 < a_2 \)). The distance between the two positions is

[MHT-CET 2024, May 11, Shift 1]

(1) \( \frac{V^2 - u^2}{a_1 - a_2} \)
(2) \( \frac{V^2 + u^2}{a_1 + a_2} \)
(3) \( \frac{u^2 + V^2}{a_1 - a_2} \)
(4) \( \frac{u^2 - V^2}{a_1 + a_2} \)

 

14. particle performs linear S.H.M. At a particular instant, the velocity of the particle is ‘u’ and the acceleration is ‘α’, while at another instant the velocity is ‘v’ and the acceleration is ‘β‘. (\( 0 < {α} < {β} \)). The distance between the two positions is

[MHT-CET 2024, May 16, Shift 1]

(1) \( \frac{u^2 - v^2}{α + β} \)
(2) \( \frac{u^2 + v^2}{α + β} \)
(3) \( \frac{u^2 - v^2}{α - β} \)
(4) \( \frac{u^2 + v^2}{α - β} \)

 

15. A particle starts oscillating simple harmonically from its equilibrium position with time period ‘T’. What is the ratio of potential energy to kinetic energy of the particle at time t = \(\frac{T}{12} \) ? (\( sin(\frac{\pi}{6}) = \frac{1}{2} \))

[MHT-CET 2024, May 10, Shift 2]

(1) 1 : 2
(2) 2 : 1
(3) 1 : 3
(4) 3 : 1

 

16. In S.H.M., the displacement of a particle at an instant is Y = A cos 30º, where A = 40 cm and kinetic energy is 200 \( J \). If force constant is \(1 × 10^x N/m\), then x will be (cos30º = \( \sqrt{3}/2 \) ) 

[MHT-CET 2024, May 9, Shift 1]

(1) 4
(2) 3
(3) 2
(4) 1

 

17. The kinetic energy of a particle, executing simple harmonic motion is 16 \( J \) when it is in mean position. If amplitude of motion is 25 cm and the mass of the particle is 5.12 kg, the period of oscillation is

[MHT-CET 2024, May 16, Shift 2]

(1) \( \frac{π}{5} \)s
(2) 2π s
(3) 20π s
(4) 5π s

 

18. The period of a simple pendulum gets doubled when

[MHT-CET 2024, May 3, Shift 2]

(1) its length is doubled
(2) its length is made four times
(3) its length is made half
(4) the mass of the bob is doubled

 

19. A simple pendulum of length ‘l’ has a brass bob attached at its lower end. It’s period is ‘T’. A steel bob of the same size, having density ‘x’ times that of brass, replaces the brass bob. Its length is then so changed that the period becomes ‘2T’. What is the new length ?

[MHT-CET 2024, May 10, Shift 2]

(1) 4lx
(2) \(\frac{4l}{x} \)
(3) 4l
(4) 2l

 

20. A simple pendulum of length \( l_1 \) has time period \( T_1 \). Another simple pendulum of length \( l_2 \)( \( l_1 > l_2 \) ) has time period \( T_2 \) . Then the time period of the pendulum of length ( \( l_1 - l_2 \) ) will be

[MHT-CET 2024, May 4, Shift 1]

(1) \( T_1 - T_2 \)
(2) \( \sqrt{\frac{T_1}{T_2}} \)
(3) \( \sqrt{\ T_1^2 - T_2^2} \)
(4) \( \sqrt{\frac{T_2}{T_1}} \)

 

21. A pendulum is oscillating with frequency ‘\( n \)’ on the surface of the Earth. If it is taken to a depth \( \frac{R}{4} \) below the surface of the Earth, the new frequency of oscillationn of depth \( \frac{R}{4} \) is (R = radius of earth)

[MHT-CET 2024, May 4, Shift 2]

(1) \( \frac{2}{\sqrt{3} n} \)
(2) \( \frac{\sqrt{3} n}{2} \)
(3) \( \frac{2 n}{\sqrt{3}} \)
(4) \( \frac{n}{4} \)

 

22. A simple pendulum has a periodic time ‘\( T_1 \)’ when it is on the surface of the Earth of radius ‘\( R \)’. Its periodic time is ‘\( T_2 \)’ when it is taken to a height ‘\( R \)’ above the earth’s surface. The value of \( \frac{T_2}{T_1} \) is

[MHT-CET 2024, May 10, Shift 2]

(1) \( \sqrt{2} \)
(2) 1
(3) 2
(4) \( \frac{1}{2} \)

 

23. Let ‘\( l_1 \)’ be the length of a simple pendulum. Its length changes to ‘\( l_2 \)’ to increase the periodic time by 20%. The ratio \( \frac{l_2}{l_1} \) =

[MHT-CET 2024, May 10, Shift 1]

(1) 1.22
(2) 1.33
(3) 1.44
(4) 1.55

 

24. Choose the correct answer,

When the point of suspension of pendulum is moved vertically upward with acceleration ‘a’, its period of oscillation

[MHT-CET 2024, May 2, Shift 1]

(1) decreases
(2) increases
(3) remains same
(4) some times increases and some times decreases

 

25. A simple pendulum of length \( L \) has mass \( m \), and it oscillates freely with amplitude \( A \). At the extreme position, its potential energy is (\( g = \text{acceleration due to gravity} \))

[MHT-CET 2024, May 16, Shift 1] [MHT-CET 2024, May 11, Shift 1]

(1) \( \frac{mgA}{2L} \)
(2) \( \frac{mgA^2}{L} \)
(3) \( \frac{mgA}{L} \)
(4) \( \frac{mgA^2}{2L} \)

 

26. A mass ‘\( m \)’ attached to a spring oscillates with a period of 3 seconds. If the mass is increased by 0.6 kg, the period increases by 3 seconds. The initial mass ‘\( m \)’ is equal to

[MHT-CET 2024, May 4, Shift 2]

(1) 0.1 kg
(2) 0.2 kg
(3) 0.3 kg
(4) 0.4 kg

 

27. Two bodies \( A \) and \( B \) of equal mass are suspended from two separate massless springs of spring constants \( K_1 \) and \( K_2 \), respectively. The two bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitude of \( B \) to that of \( A \) is

[MHT-CET 2024, May 4, Shift 1]

(1) \( \frac{K_1}{K_2} \)
(2) \( \frac{K_2}{K_1} \)
(3) \( \sqrt{\frac{K_1}{K_2}} \)
(4) \( \sqrt{\frac{K_2}{K_1}} \)

 

28. A spring has a certain mass suspended from it and its period of vertical oscillations is \( T_1 \). The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now \( T_2 \). The ratio of \( \frac{T_2}{T_1} \) is

[MHT-CET 2024, May 2, Shift 2]

(1) 1 : 2
(2) 1: \( \sqrt{2} \)
(3) \( \sqrt{2} \) : 1
(4) 2 : 1

 

29. A spring has length \( L \) and force constant \( K \). It is cut into two springs of lengths \( L_1 \) and \( L_2 \) such that \( L_1 = N L_2 \) (\( N \) is an integer). The force constant of the spring of length \( L_1 \) is

[MHT-CET 2024, May 15, Shift 2]

(1) (N + 1)K
(2) \( \frac{K}{N}(1 + N) \)
(3) \( K \)
(4) \( \frac{K}{N + 1} \)

 

30. All the springs in fig. (a), (b), and (c) are identical, each having a force constant \( K \). Mass attached to each system is ‘\( m \)’. If \( T_a \), \( T_b \), and \( T_c \) are the time periods of oscillations of the three systems respectively, then

[MHT-CET 2024, May 11, Shift 2]

(1) \( T_a = \sqrt{2}T_b \)
(2) \( T_a = \frac{T_c}{\sqrt{2}} \)
(3) \( T_b = 2T_a \)
(4) \( T_b = 2T_c \)

 

31. Three masses 500 g, 300 g and 100 g are suspended at the end of spring as shown in figure and are in the equilibrium. When the 500 g mass is removed, the system oscillates with a period of 3 second. When the 300 g mass is also removed it will oscillate with a period of

[MHT-CET 2024, May 2, Shift 1]

(1) 1 s
(2) 1.5 s
(3) 2 s
(4) 2.5 s

 

32. 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

[MHT-CET 2024, May 9, Shift 1]

(1) 3 U
(2) 4 U
(3) 5 U
(4) 9 U

 

33. The 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. The maximum speed of the particle is (g = 10, m/ \( s^2 \))

[MHT-CET 2024, May 3, Shift 2]

(1) \( \frac{1}{\pi} \) m/s
(2) \( \frac{1}{2\pi} \) m/s
(3) \( \frac{1}{4\pi} \) m/s
(4) \( 2\pi \) m/s

 

34. A piece of wood has length, breadth, and height ‘\( a \)’, ‘\( b \)’, and ‘\( c \)’, respectively. Its relative density is ‘\( d \)’. It is floating in water such that the side ‘\( a \)’ is vertical. It is pushed down a little and released. The time period of the S.H.M. executed by it is (\( g = \text{acceleration due to gravity} \))

[MHT-CET 2024, May 11, Shift 2]

(1) \( 2\pi \sqrt{\frac{abc}{g}} \)
(2) \( 2\pi \sqrt{\frac{bc}{dg}} \)
(3) \( 2\pi \sqrt{\frac{g}{da}} \)
(4) \( 2\pi \sqrt{\frac{ad}{g}} \)

 

35. A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is 1.6m. The period of oscillation of the sphere in seconds is (acceleration due to gravity, (g = 10, m/ \( s^2 \))

[MHT-CET 2024, May 3, Shift 1]

(1) 0.8 π
(2) 0.6 π
(3) 0.4 π
(4) 0.2 π

 

36. A tube of uniform bore of cross-sectional area ‘\( A \)’ has been set up vertically with its open end facing up. Now ‘\( M \)’ gram of a liquid of density ‘\( d \)’ is poured into it. The column of liquid in this tube will oscillate with a period ‘\( T \)’, which is equal to [\( g = \text{acceleration due to gravity} \)]

[MHT-CET 2024, May 3, Shift 1]

(1) \( 2\pi \sqrt{\frac{MA}{gd}} \)
(2) \( 2\pi \sqrt{\frac{M}{2Adg}} \)
(3) \( 2\pi \sqrt{\frac{M}{g}} \)
(4) \( 2\pi \sqrt{\frac{M}{gdA}} \)

 

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

[MHT-CET 2024, May 2, Shift 1] 

(1) \( \sqrt{\frac{k}{m} + (\frac{b}{2m})^2} \)
(2) \( \frac{k}{m} + (\frac{b}{2m})^2 \)
(3) \( \sqrt{\frac{k}{m} - (\frac{b}{2m})^2} \)
(4) \( \frac{k}{m} - (\frac{b}{2m})^2 \)

 

We will post the solution to this paper in 24 hours. The download link will be activated at that time. In the meantime, students are encouraged to attempt the questions on their own and use the comments section to discuss and collaborate with each other.

 

Download solution here.

 

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