JEE PHYSICS MCQ: OSCILLATIONS

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A particle which is attached to a spring oscillates horizontally with simple harmonic motion with a frequency of

1 / \pi \mathrm{~Hz}

and total energy of

10 \mathrm{~J}

. If the maximum speed of the particle is

0.4 \mathrm{~ms}^{-1}

, what is the force constant of the spring? What will be the maximum potential energy of the spring during the motion?

View answer and explanation

Correct option: A

500 \mathrm{~Nm}^{-1}, 10 \mathrm{~J}

Option A (

500 \text{ Nm}^{-1}, 10 \text{ J}

) is correct because the force constant and maximum potential energy are calculated correctly using the given frequency, total energy, and maximum speed.

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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)