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

Answer: (D)

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