# A simple harmonic oscillator has an amplitude $\alpha$ and time period T. The time required by it to travel from $x = \alpha$  to  $x = \frac{\alpha}{2}$  is Option 1) $4.0\: s$ Option 2) $2.0\: s$ Option 3) $1.0\: s$ Option 4) $0.5\: s$

A Avinash

It is required to calculate the time from extreme position. Hence, in this case equation for displacement of particle can be written as $\\*x = a\sin(\omega t + \frac{\pi}{2}) = a\cos\omega t \\*\Rightarrow \frac{a}{2} = a\cos \omega t \Rightarrow \omega t = \frac{\pi}{3} \Rightarrow \frac{2\pi}{T}\cdot t = \frac{\pi}{3} \Rightarrow t = \frac{T}{6}$

Time Period -

Since all periodic motion repeat themselves in equal time interval. This minimum time interval is known as time period for oscillation.

- wherein

It is denoted by T.

Option 1)

$4.0\: s$

This is correct.

Option 2)

$2.0\: s$

This is incorrect.

Option 3)

$1.0\: s$

This is incorrect.

Option 4)

$0.5\: s$

This is incorrect.

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