The maximum velocity of a particle,executing | vedclass

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(B) The maximum velocity of a particle in simple harmonic motion is given by $v_{\max} = a \omega$,where $a$ is the amplitude and $\omega$ is the angu

Vedclass · Accepted Answer

$0.01$

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(B) The maximum velocity of a particle in simple harmonic motion is given by $v_{\max} = a \omega$,where $a$ is the amplitude and $\omega$ is the angular frequency.Since $\omega = \frac{2 \pi}{T}$,we have $v_{\max} = a 	imes \frac{2 \pi}{T}$.Rearranging for the period $T$,we get $T = \frac{2 \pi a}{v_{\max}}$.Given $a = 7 \ mm = 7 	imes 10^{-3} \ m$ and $v_{\max} = 4.4 \ m/s$.Substituting the values: $T = \frac{2 	imes 3.14 	imes 7 	imes 10^{-3}}{4.4}$.$T = \frac{43.96 	imes 10^{-3}}{4.4} \approx 9.99 	imes 10^{-3} \ s \approx 0.01 \ s$.

The maximum velocity of a particle,executing simple harmonic motion with an amplitude $7 \ mm$,is $4.4 \ m/s$. The period of oscillation is .... $sec$

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