Physics - SHM Question with Solution | TestHub

Let the amplitude of the S.H.M. be $A$ and the angular frequency be $\omega$.

The maximum velocity is given by $v_{max}=A\omega =\alpha$.

The maximum acceleration is given by $a_{max}=A{\omega }^2=\beta$.

Squaring the maximum velocity, we get $A^2{\omega }^2={\alpha }^2$.

Dividing this by the maximum acceleration, we get:

$\frac{A^2{\omega }^2}{A{\omega }^2}=\frac{{\alpha }^2}{\beta }$

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

The path length of a particle in linear S.H.M. is the distance between the two extreme positions, which is $2A$.

Path length $=2A=\frac{2{\alpha }^2}{\beta }$

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