Question 283: Oscillations

Question

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

Accepted Answer

For S.H.M.,
Maximum velocity, $\alpha=\mathrm{A} \omega$
$$\omega=\frac{\alpha}{A}\quad	ext{.... (i)}$$
Maximum acceleration, $\beta=\mathrm{A} \omega^2$
$$\begin{aligned}
& \beta=A\left(\frac{\alpha}{A}ight)^2=\frac{\alpha^2}{A} \
& \Rightarrow A=\frac{\alpha^2}{\beta}
\end{aligned}$$
$	herefore\quad$ Path length $=2 A=\frac{2 \alpha^2}{\beta}$

Answer

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

Answer

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

Answer

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

Answer

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

Oscillations

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

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