First slide

Simple hormonic motion

Question

A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency $ω$ . The amplitude of oscillation is gradually increased. The
coin will leave contact with the platform for the first time

Moderate

A

at the mean position of the platform
B

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

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

at the highest position of the platform

Solution

The net effect of these two forces must be towards mean position.
At the mean position, there is no net force and hence normal reaction equals mg. Above mean position, normal reaction is less than mg and below mean position, normal reaction is greater than mg.
At the top extreme:
N is minimum at the top extreme.
Net force towards mean position = ma = ${mω}^{2} A$
$\Rightarrow mg - N_{t} = {mω}^{2} A$
$N_{t} = N_{\min} = mg - {mω}^{2} A$
For loosing contact with the platform, N= 0
So, A = $\frac{g}{ω^{2}}$

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