The displacement of a particle executing S.H. | vedclass

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(B) The given equation for displacement is $x = 0.01 \sin 100 \pi(t + 0.05)$.Comparing this with the standard $S.H.M.$ equation $x = A \sin(\omega t +

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$0.02$

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(B) The given equation for displacement is $x = 0.01 \sin 100 \pi(t + 0.05)$.Comparing this with the standard $S.H.M.$ equation $x = A \sin(\omega t + \phi)$,we identify the angular frequency $\omega = 100 \pi \, rad/s$.The time period $T$ is related to the angular frequency by the formula $T = \frac{2 \pi}{\omega}$.Substituting the value of $\omega$:$T = \frac{2 \pi}{100 \pi} = \frac{1}{50} = 0.02 \, s$.Thus,the time period is $0.02 \, s$.

The displacement of a particle executing $S.H.M.$ is given by $x = 0.01 \sin 100 \pi(t + 0.05)$. The time period is ........ $s$.

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