Question

The maximum acceleration of the body executing SHM is $a_{0}$ and maximum velocity is $v_{0}$ the amplitude is given by

• A. $\frac{1}{a_{0} v_{0}}$
• B. $\frac{a_{0}^{2}}{v_{0}}$
• C. $v_{0} a_{0}$
• D. $\frac{v_{0}^{2}}{a_{0}}$

Answer 1

Answer: (D)

For a body executing SHM, maximum acceleration $\left(a_{0}\right)$ is given by
$\omega^{2} A=a_{0}$ ...(i)
and maximum velocity $\left(v_{0}\right)$ is
$A \omega=v_{0}$ ...(ii)
where $A$ is amplitude, $\omega$ the angular velocity.
Squaring Eq. (ii) and dividing by Eq. (i), we get
$A=\frac{v_{0}^{2}}{a_{0}}$