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(A) The given equation for displacement is $y = A \sin(\omega t + \phi)$,where $A = 0.30 \, m$ and $\omega = 220 \, rad/s$.$1$. Frequency $(n)$: The f

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$35 \, Hz$,$66 \, m/s$

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(A) The given equation for displacement is $y = A \sin(\omega t + \phi)$,where $A = 0.30 \, m$ and $\omega = 220 \, rad/s$.$1$. Frequency $(n)$: The frequency is given by $n = \frac{\omega}{2\pi}$.$n = \frac{220}{2 	imes 3.14159} \approx 35.01 \, Hz \approx 35 \, Hz$.$2$. Maximum velocity $(v_{\max})$: The maximum velocity is given by $v_{\max} = \omega A$.$v_{\max} = 220 	imes 0.30 = 66 \, m/s$.Thus,the frequency is $35 \, Hz$ and the maximum velocity is $66 \, m/s$.

If the displacement of a particle executing $SHM$ is given by $y = 0.30 \sin(220t + 0.64)$ in meters,then the frequency and maximum velocity of the particle are:

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