First slide

Simple hormonic motion

Question

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Moderate

A

$\frac{\sqrt{5}}{2 π}$
B

$\frac{4 π}{\sqrt{5}}$
C

$\frac{2 π}{\sqrt{3}}$
D

$\frac{\sqrt{5}}{π}$

Solution

Amplitude A = 3 cm
When particle is at x = 2 cm,
its |velocity| = |acceleration|
i.e., $ω \sqrt{A^{2} - x^{2}} = ω^{2} x \Rightarrow ω = \frac{\sqrt{A^{2} - x^{2}}}{x}$
$T = \frac{2 π}{ω} = 2 π (\frac{2}{\sqrt{5}}) = \frac{4 π}{\sqrt{5}}$

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