First slide

Simple harmonic motion

Question

A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 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

$\frac{1}{2 π \sqrt{3}}$
B

$2 π \sqrt{3}$
C

$\frac{2 π}{\sqrt{3}}$
D

$\frac{\sqrt{3}}{2 π}$

Solution

Velocity $v = ω \sqrt{A^{2} - x^{2}}$ and acceleration $= ω^{2} x$
Now given, $ω^{2} x = ω \sqrt{A^{2} - x^{2}}$ $\Rightarrow ω^{2} . 1 = ω \sqrt{2^{2} - 1^{2}}$
$\Rightarrow ω = \sqrt{3}$
$∴ T = \frac{2 π}{ω} = \frac{2 π}{\sqrt{3}}$

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