A) 1
B) 0.75
C) 0.5
D) 0.25
Correct Answer: D
Solution :
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 Eq. (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
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