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A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency  \omega 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     \frac{g}{\omega ^{2}}

  • Option 4)

    or an amplitude of   \frac{g^{2}}{\omega ^{2}}

 

Answers (1)

best_answer

As we learnt in

Equation of S.H.M. -

a=-\frac{d^{2}x}{dt^{2}}= -w^{2}x

w= \sqrt{\frac{k}{m}}

 

- wherein

x= A\sin \left ( wt+\delta \right )

 

 

 

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     \frac{g}{\omega ^{2}}

This is the correct option.

Option 4)

or an amplitude of   \frac{g^{2}}{\omega ^{2}}

This is an incorrect option.

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