JEE PHYSICS MCQ: OSCILLATIONS

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The period of oscillation of a simple pendulum is given by

T=2 \pi \sqrt{\frac{L}{g}}

, where

L

is the length of the pendulum and

g

is the acceleration due to gravity. The length is measured using a meter scale which has 2000 divisions. If the measured value

L

50 \mathrm{~cm}

, the accuracy in the determination of

g

1.1 \%

and the time taken for 100 oscillations is 100 seconds, what should be the resolution of the clock (in milliseconds)?

View answer and explanation

Correct option: C

Option C (5) is correct because the resolution of the clock is calculated by error propagation formula from the given percentage error in g and the least count of the meter scale.

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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)