Obtain the ratio of maximum acceleration and maximum velocity of a $SHM$ particle.
(D) For a particle executing $SHM$ with amplitude $A$ and angular frequency $\omega$,the displacement is given by $x(t) = A \sin(\omega t + \phi)$.
The velocity is $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$. The maximum velocity is $v_{max} = A\omega$.
The acceleration is $a(t) = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi)$. The maximum acceleration is $a_{max} = A\omega^2$.
The ratio of maximum acceleration to maximum velocity is $\frac{a_{max}}{v_{max}} = \frac{A\omega^2}{A\omega} = \omega$.
The velocity is $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$. The maximum velocity is $v_{max} = A\omega$.
The acceleration is $a(t) = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi)$. The maximum acceleration is $a_{max} = A\omega^2$.
The ratio of maximum acceleration to maximum velocity is $\frac{a_{max}}{v_{max}} = \frac{A\omega^2}{A\omega} = \omega$.
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