The maximum speed of a particle executing S.H | vedclass

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(D) Given: Maximum velocity $v_{max} = \omega A = 1 \ m/s$ and maximum acceleration $a_{max} = \omega^2 A = 1.57 \ m/s^2$.Dividing the expression for

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$4$

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(D) Given: Maximum velocity $v_{max} = \omega A = 1 \ m/s$ and maximum acceleration $a_{max} = \omega^2 A = 1.57 \ m/s^2$.Dividing the expression for maximum acceleration by the expression for maximum velocity:$\frac{a_{max}}{v_{max}} = \frac{\omega^2 A}{\omega A} = \omega$$\omega = \frac{1.57}{1} = 1.57 \ rad/s$.We know that $\omega = \frac{2\pi}{T}$.Substituting $\pi \approx 3.14$:$1.57 = \frac{2 	imes 3.14}{T}$$T = \frac{6.28}{1.57} = 4 \ s$.Therefore,the time period of the particle is $4 \ s$.

The maximum speed of a particle executing $S.H.M.$ is $1 \ m/s$ and its maximum acceleration is $1.57 \ m/s^2$. The time period of the particle will be .... $s$.

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