Question

A body executing SHM has a maximum velocity of $1m{s}^{-1}$ and a maximum acceleration of $4m{s}^{-2}$ . Its amplitude in metre is :

• A. 1
• B. 0.75
• C. 0.5
• D. 0.25

Answer 1

The correct option is D 0.25

A body executing SHM have $V_{max}=1m{s}^{-1}$
and $a_{max}=4m{s}^{-1}$
Since, $v_{max}=r\omega$ ..(i)
and $a_{max}={\omega }^{2}r$ ...(ii)
Where, $\omega =$ angular speed of SHM
r = amplitude of SHM
Dividing Eq. (ii) by the square of Eq. (i), we get
$\frac{a_{max}}{v_{max}^2}=\frac{{\omega }^{2}r}{r^2{\omega }^2}=\frac{1}{r}$
$r=\frac{v_{max}^2}{a_{max}}=\frac{1\times 1}{4}=\frac{1}{4}$
r = 0.25 m
$\therefore$ Amplitude of SHM = 0.25 m