JEE PHYSICS MCQ: OSCILLATIONS

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Time period of a simple pendulum of length

\ell

T_1

and time period of uniform rod of same length

\ell

pivoted about one end and oscillating in a vertical plane is

T_2

. Amplitude of oscillations in both the cases is small, then

\frac{T_1}{T_2}

View answer and explanation

Correct option: C

\sqrt{\frac{3}{2}}

Option C (

\sqrt{\frac{3}{2}}

) is correct because it represents the ratio of the time periods of a simple pendulum and a uniform rod oscillating about one end, calculated using their respective formulas.

More Questions from OSCILLATIONS

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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)