The maximum velocity and maximum acceleration of a particle performing a linear S.H.M. is ' $\alpha$ ' and '…
The maximum velocity and maximum acceleration of a particle performing a linear S.H.M. is ' $\alpha$ ' and ' $\beta$ ' respectively. Then the path length of the particle is
$\frac{\alpha^2}{\beta}$
$\frac{\beta \alpha^2}{2 \alpha^2}$
$\frac{2 \alpha^2}{\beta}$
$\frac{2 \beta}{\alpha^2}$
Solution
For S.H.M.,
Maximum velocity, $\alpha=\mathrm{A} \omega$
$\omega=\frac{\alpha}{\mathrm{A}}$...(i)
Maximum acceleration, $\beta=A \omega^2$
$\begin{aligned}
& \beta=A\left(\frac{\alpha}{A}\right)^2=\frac{\alpha^2}{A} \\
& \Rightarrow A=\frac{\alpha^2}{\beta}
\end{aligned}$
$\therefore \quad$ Path length $=2 \mathrm{~A}=\frac{2 \alpha^2}{\beta}$