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Question
A particle moves on the X-axis according to the equation x = x0 sin2 ωt. The motion is simple harmonic
Options
with amplitude x0
with amplitude 2x0
with time period \[\frac{2\pi}{\omega}\]
with time period \[\frac{\pi}{\omega} .\]
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Solution
with time period \[\frac{\pi}{\omega} \]
Given equation:
x = xo sin2 ωt
⇒ \[x = \frac{x_0}{2}(\cos 2\omega t - 1)\]
Now, the amplitude of the particle is xo/2 and the angular frequency of the SHM is 2ω.
Thus, time period of the SHM =
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