Test Index

Oscillations Part 1

Section: Physics

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Question : 81 of 100

Marks: +1, -0

A coin is placed on a horizontal platform which undergoes vertical simple harmonic motoin of angular frequency

\omega

. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

at the mean position of the platform
for an amplitude of $\frac{g}{\omega^2}$
For an amplitude of $\frac{g^2}{\omega^2}$
at the height position of the platform

Solution:

Coin

A

is moving in simple harmonic motion in vertical direction. Now we are assuming coin will leave contact with the platform when platform is at a distance of

x

from the mean position which is also called amplitude.

At distance

x

the force acting on the coin is

m g - N = m \omega^2 x

For coin to leave contact

N=0

\Rightarrow m g = m \omega^2 x \Rightarrow x = \frac{g}{\omega^2}

\therefore

Option (B) is correct.

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