Question

The time period of oscillation of a particle that executes S.H.M. is 1.2 sec.The time starting from extreme position,its velocity will be half of its velocity at mean position is​

Answer 1

Explanation:

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Answer 2

The time taken is 0.1 s.

Explanation:

The time period of oscillation is T = 1.2 s.

Time is zero at extreme position.

The velocity of the particle executing S.H.M is given as

 $v=\omega\sqrt{A^2-x^2}$

Here A is the amplitude, x is the displacement and  $\omega$ is the angular frequency.

Also the maximum velocity in S.H.M is given as

$v_{max}=\omega A$

As per question,

$v=\frac{v_{max} }{2}$

so,

$\omega\sqrt{A^2-x^2}=\frac{\omega A}{2}$

$\frac{A^2}{4} =(A^2-x^2)$

$x=\frac{\sqrt{3} }{2}A$

Equation of S.H.M is given as

$x=Acos\omega t$

Equating x from above equation, we get

$\frac{\sqrt{3} }{2}A = Acos\omega t$

$t=\frac{\pi}{6\times\omega}$

$t=\frac{\pi}{6\times\frac{2\pi}{T} }$               ( $\because$  $\omega=\frac{2\pi}{T}$)

substitute the value of T, we get

$t=\frac{\pi}{6\times\frac{2\pi}{1.2s} }$

t = 0.1 s.

Thus,  the time taken from extreme position to mean position is 0.1 s.

#Learn More: Velocity in S.H.M.

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