A coin is placed on a horizontal platform which undergoes vertica… - Vidyadip

MCQ

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

A
at the mean position of the platform
✓
for an amplitude of $\frac{g}{\omega ^2}$
C
for an amplitude of $\frac{g^2}{\omega ^2}$
D
at the highest position of the platform

✓

Answer

Correct option: B.

for an amplitude of $\frac{g}{\omega ^2}$

b
For block $A$ to move in $SHM.$

$\mathrm{mg}-\mathrm{N}=\mathrm{m} \omega^{2} \mathrm{x}$

where $x$ is the distance from mean position

For block to leave contact $\mathrm{N}=0$

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

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