A particle is executing a simple harmonic mot | vedclass

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(A) For a particle in simple harmonic motion with amplitude $A$ and angular frequency $\omega$:Maximum acceleration is given by $\alpha = A\omega^2$.M

Vedclass · Accepted Answer

$\frac{2\pi \beta}{\alpha}$

Vedclass · Answer

(A) For a particle in simple harmonic motion with amplitude $A$ and angular frequency $\omega$:Maximum acceleration is given by $\alpha = A\omega^2$.Maximum velocity is given by $\beta = A\omega$.Dividing the two equations: $\frac{\alpha}{\beta} = \frac{A\omega^2}{A\omega} = \omega$.Therefore,$\omega = \frac{\alpha}{\beta}$.The time period $T$ is given by $T = \frac{2\pi}{\omega}$.Substituting the value of $\omega$: $T = \frac{2\pi}{(\alpha / \beta)} = \frac{2\pi \beta}{\alpha}$.

$A$ particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$. Then,its time period of vibration will be

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