For a simple harmonic motion given by x = 0.0 | vedclass

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(B) The standard equation for simple harmonic motion is given by $x = A \sin(\omega t + \phi)$.Comparing the given equation $x = 0.01 \sin(100\pi t +

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$0.02$

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(B) The standard equation for simple harmonic motion is given by $x = A \sin(\omega t + \phi)$.Comparing the given equation $x = 0.01 \sin(100\pi t + 5\pi)$ with the standard equation,we identify the angular frequency $\omega = 100\pi 	ext{ rad/s}$.The time period $T$ is related to the angular frequency by the formula $T = \frac{2\pi}{\omega}$.Substituting the value of $\omega$,we get $T = \frac{2\pi}{100\pi} = \frac{2}{100} = 0.02 	ext{ sec}$.Therefore,the time period of the oscillation is $0.02 	ext{ sec}$.

For a simple harmonic motion given by $x = 0.01 \sin 100\pi (t + 0.05)$,what is the time period in $\text{sec}$?

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