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Question

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

Vedclass · Accepted Answer

$60\pi 	ext{ cm/s}$

Vedclass · Answer

(A) The maximum velocity $(v_{\max})$ of a particle in simple harmonic motion is given by the formula $v_{\max} = a\omega$,where $a$ is the amplitude and $\omega$ is the angular frequency.Given:Amplitude $a = 0.1 	ext{ cm}$Frequency $n = 300 	ext{ Hz}$Angular frequency $\omega = 2\pi n = 2 	imes \pi 	imes 300 = 600\pi 	ext{ rad/s}$.Substituting these values into the formula:$v_{\max} = 0.1 	imes 600\pi = 60\pi 	ext{ cm/s}$.Therefore,the maximum velocity of the particle is $60\pi 	ext{ cm/s}$.

When a sound wave of frequency $300 Hz$ passes through a medium,the maximum displacement of a particle of the medium is $0.1 cm$. The maximum velocity of the particle is equal to:

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