JEE PHYSICS MCQ: OSCILLATIONS

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A uniform disc of radius

r

is to be suspended through a small hole made in the disc. Find the minimum possible time period of the disc for small oscillations. What should be the distance of the hole from the centre for it to have minimum time period?

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Correct option: A

2 \pi \sqrt{\frac{r \sqrt{2}}{g}}, r / \sqrt{2}

Option A (

2 \pi \sqrt{\frac{{r \sqrt{{2}}}}{{g}}}, r / \sqrt{{2}}

) is correct because it represents the minimum time period calculated using the moment of inertia and the condition for minimum time period.

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)