A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency  The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time Option 1) at the highest position of the platform Option 2) at the mean position of the platform Option 3) for an amplitude of     Option 4) or an amplitude of

As we learnt in

Equation of S.H.M. -

- wherein

In vertical simple harmonic motion, maximum acceleration$\left ( a\omega ^{2} \right )$        and so the maximum force$\left ( ma\omega ^{2} \right )$        will be at extreme positions. At highest position, force will be towards mean position and so it will be downwards. At lowest position, force will be towards mean position and so it will be upwards This is opposite to weight direction of the coin. The coin will leave contact will the platform for the first time when $m\left ( a\omega ^{2} \right )\geq mg$      at the lowest position of the platform.

Correct  option is 3.

Option 1)

at the highest position of the platform

This is an incorrect option.

Option 2)

at the mean position of the platform

This is an incorrect option.

Option 3)

for an amplitude of

This is the correct option.

Option 4)

or an amplitude of

This is an incorrect option.

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE