Question

If the maximum acceleration of the body executing S.H.M. is a${}_0$ and maximum velocity is V${}_0$ the amplitude is given by

• A. $\frac{1}{a_0{v}_0}$
• B. $\frac{a_0}{v_0}$
• C. $v_0{a}_0$
• D. $\frac{v_0^2}{a_0}$

Answer 1

The correct option is D $\frac{v_0^2}{a_0}$

Max. acceleration in S.H.M, $a_0={\omega }^{2}a$ .....(i) [a = amplitude]
max. velocity = $\omega$a
$v_0=\omega a$
$\Rightarrow v_0^2={\omega }^2{a}^2$ ........ (ii)
$\therefore \frac{\left(ii\right)}{\left(i\right)}\Rightarrow a=\frac{v_0^2}{a_0}$