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Q.

A particle moves on the x-axis according to the equation x = x₀ sin 2ωt. The motion is simple harmonic.

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a

With time period $\frac{π}{ω}$

b

With amplitude 2x_o

c

With time period 2π/ω

d

With amplitude x_o/2

answer is C.

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Detailed Solution

$x = {x_0}\sin 2\omega t$

At x = x₀,

x₀ = x₀ sin 2ωt

$\Rightarrow \sin 2\omega t = 1$

$\Rightarrow 2\omega t = \frac{\pi }{2},\left( {\pi - \frac{\pi }{2}} \right),\left( {2\pi + \frac{\pi }{2}} \right),\left( {3\pi - \frac{\pi }{2}} \right)....$

$\therefore t = \frac{\pi }{{4\omega }},\frac{{5\pi }}{{4\omega }},\frac{{9\pi }}{{4\omega }},\frac{{11\pi }}{{4\omega }},....$

$\therefore T = \left( {\frac{{5\pi }}{{4\omega }} - \frac{\pi }{{4\omega }}} \right) = \left( {\frac{{9\pi }}{{4\omega }} - \frac{{5\pi }}{{4\omega }}} \right) = \left( {\frac{{11\pi }}{{4\omega }} - \frac{{9\pi }}{{4\omega }}} \right) = ....$

$\Rightarrow T = \frac{\pi }{\omega }.$

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