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 [2017]

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Solution

Q.
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 [2017]

1 $\frac{\sqrt{5}}{2 π}$

2 $\frac{4 π}{\sqrt{5}}$

3 $\frac{2 π}{\sqrt{3}}$

4 $\frac{\sqrt{5}}{π}$

Ans.
(2)

Given, $A = 3$ cm, $x = 2$ cm

The velocity of a particle in simple harmonic motion is given as $v = ω \sqrt{A^{2} - x^{2}}$

and magnitude of its acceleration is $a = ω^{2} x$

Given $| v |$ $=$ $| a |$ $∴$ $ω \sqrt{A^{2} - x^{2}} = ω^{2} x$

$ω x = \sqrt{A^{2} - x^{2}} or ω^{2} x^{2} = A^{2} - x^{2}$

$ω^{2} = \frac{A^{2} - x^{2}}{x^{2}} = \frac{9 - 4}{4} = \frac{5}{4} or ω = \frac{\sqrt{5}}{2}$

Time period, $T = \frac{2 π}{ω} = 2 π . \frac{2}{\sqrt{5}} = \frac{4 π}{\sqrt{5}}$ $s$

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