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)

0.25

Answer 2

A body executing $\text{SHM}$ have ${{\text{V}}_{\max }}=1m{{s}^{-1}}$ and ${{a}_{\max }}=4m{{s}^{-1}}$ Since, ${{V}_{\max }}=r\omega$ ?(i) and ${{a}_{\max }}={{\omega }^{2}}x$ ?(ii) (numerically)where $\omega =$ angular speed of SHM $r=$ amplitude of SHM Dividing E (ii) by the square of(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