Back to OSCILLATIONS

JEE PHYSICS MCQ: OSCILLATIONS

1 Medium

1 of 1

Application +4 | -1

A particle performs Simple Harmonic Motion of amplitude

{A}

and angular frequency

\omega

along a straight line. When it is at a distance

\frac{\sqrt{3}}{2} {~A}

from mean position, its kinetic energy gets increased suddenly by an amount

\frac{1}{2} m \omega^2 A^2

, due to an impulsive force. Then new amplitude of oscillations of the particle will be :

View answer and explanation

Correct option: C

\sqrt{2} {~A}

Option C (

\sqrt{2} {~A}

) is correct because the new amplitude is calculated by equating the new total energy after the impulse to the potential energy at the new amplitude.

More Questions from OSCILLATIONS

Practice All Chapter MCQs →

Assertion: A particle executing simple harmonic motion comes to rest at the extreme positions.

Reason: The resultant force on the particle is zero at these positions. A particle is subjected simultaneously to two SHM’s one along the

x

-axis and the other along the

y

-axis. The two vibrations are in phase and have unequal amplitudes. The particle will execute A particle is executing SHM with amplitude

a

. When the PE of a particle is one-fourth of its maximum value during the oscillation, its displacement from the equilibrium position will be The displacement, velocity and acceleration in a simple harmonic motion are related as the Assertion: Angular velocity of earth around sun is lesser than the angular velocity of moon about earth.
Reason: Time taken by moon revolve around earth is less than time taken by earth to revolve around sun. A particle is executing

SHM

. At a displacement

{y_1}

its potential energy is

{U_1}

and at a displacement

{y_2}

its potential energy is

{U_2}

. The potential energy of the particle at displacement

\left( {{y_1} + {y_2}} \right)