Question

A body executing SHM has a maximum velocity of $1 \, ms^{-1} $ and a maximum acceleration of $4 \, ms^{-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 $V _{\max }=1\, ms ^{-1}$
and $a _{\max }=4 \,ms ^{-1}$
Since, $v _{\max }= r \omega$..(i)
and $a _{\max }=\omega^{2} r \ldots$ (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 $=0.25\, m$