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Oscillations MCQs for NEET — Physics Questions with Answers

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If a particle is executing SHM, with an amplitude A, the distance moved and the displacement of the body in a time equal to its period are

2A, A

4A, 0

A, A

0, 2A

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Explanation

Distance = 4A

Displacement = 0

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$The equation of the displacement of two particles making SHM are represented by y_{1} = a \sin (ωt + ϕ) & y_{2} = a \cos (ωt) The phase difference of the velocities of the t w o particles is$

$\frac{π}{2} + ϕ$

$- ϕ$

$ϕ$

$ϕ - \frac{π}{2}$

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The displacement of a particle executing SHM is given by y = 0.25 (sin 200t) cm. The maximum speed of the particles is:

200 cm/sec

100 cm/sec

50 cm/sec

0.25 cm/sec

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Explanation

$Maximum particle velocity = aω = 0.25 x 200 = 50 cm / s$

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$A particle is executing SHM with amplitude A and time period T . If at t = 0, it is at origin (mean position) then find the time instant, when it covers a distance equal to \frac{5 A}{2}$

$\frac{T}{12}$

$\frac{5 T}{12}$

$\frac{7 T}{12}$

$\frac{2 T}{3}$

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A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude?

$\frac{1}{2} s$

$\frac{1}{6} s$

$\frac{1}{4} s$

$\frac{1}{3} s$

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Explanation

$y = A \sin ω t \frac{A}{2} = A \sin ω t \sin ω t = \frac{1}{2} = \sin (\frac{π}{6}) ω t = \frac{π}{6} \Rightarrow \frac{2 π}{T} t = \frac{π}{6} t = \frac{T}{12} = \frac{2}{12} = \frac{1}{6} s e c$

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If the displacement (x) and velocity v of a particle executing simple harmonic motion are related through the expression $4 v^{2} = 25 - x^{2}$ then its time period is:

$π$

2 $π$

4 $π$

6 $π$

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Explanation

$S \tan d a r d e q u a t i o n o f S H M v^{2} = ω^{2} (A^{2} - x^{2}) 4 v^{2} = 25 - x^{2} v^{2} = \frac{1}{4} (25 - x^{2}) C o m p a r e i t w i t h v^{2} = ω^{2} (A^{2} - x^{2}) \Rightarrow ω^{2} = \frac{1}{4} \Rightarrow ω = \frac{1}{2} T i m e p e r i o d = \frac{2 π}{ω} = 4 π$

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Two simple pendulums have time periods T and $\frac{5 T}{4}$ . They start vibrating at the same instant from the mean position in the same phase. The phase difference between them when bigger pendulum completes one oscillation will be:

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$

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Explanation

$ϕ = ω_{s} t - ω_{i} t ϕ = \frac{2 π}{T} . \frac{5 T}{4} - \frac{2 π}{\frac{5 T}{4}} . \frac{5 T}{4}$

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There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is T. If the resultant acceleration becomes g/4, then the new time period of the pendulum is

0.8 T

0.25 T

2 T

4 T

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Explanation

When lift is at rest, $T = 2 π \sqrt{l / g}$

If acceleration becomes g/4 then

$T^{'} = 2 π \sqrt{\frac{l}{g / 4}} = 2 π \sqrt{\frac{4 l}{g}} = 2 \times T$

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A particle executes linear simple harmonic motion with an amplitude of of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then, its time period in seconds is

$\sqrt{\frac{5}{π}}$

$\sqrt{\frac{5}{2 π}}$

$\frac{4 π}{\sqrt{5}}$

$\frac{2 π}{\sqrt{3}}$

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Explanation

= $ω \sqrt{A^{2} - X^{2}}$

Also, magnitude of acceleration of particle in SHM

= $ω^{2} x$

Given, when x=2cm

|v|=|a|

$\Rightarrow ω \sqrt{A^{2} - X^{2}} = ω^{2} X$

$\Rightarrow ω = \frac{\sqrt{A^{2} - X^{2}}}{X}$

= $\frac{\sqrt{9 - 4}}{2}$

$\Rightarrow$ Angular velocity $ω = \frac{\sqrt{5}}{2}$

So, Time period of motion

T= $\frac{2 π}{ω}$ = $\frac{4 π}{\sqrt{5}} s$

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A body mass m is attached to the lower end of a spring whose upper end is fixed. The spring has neglible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5s. The value of m in kg is-

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{16}{9}$

$\frac{9}{16}$

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