A) 1
B) 0.75
C) 0.5
D) 0.25
Correct Answer: D
Solution :
A body executing SHM have \[{{V}_{\max }}=1\,m{{s}^{-1}}\] and \[{{a}_{\max }}=4\,m{{s}^{-1}}\] since, \[{{v}_{\max }}=r\omega \] ?. (i) and \[{{a}_{\max }}={{\omega }^{2}}r\] ?. (ii) (numerically) where,\[\omega =\] angular speed of SHM r = amplitude of SHM Dividing Eq. (ii) by the square of Eq. (i), we get \[\frac{{{a}_{\max }}}{V_{\max }^{2}}=\frac{{{\omega }^{2}}r}{{{r}^{2}}{{\omega }^{2}}}=\frac{1}{r}\] \[r=\frac{v_{\max }^{2}}{{{a}_{\max }}}=\frac{1\times 1}{4}=\frac{1}{4}\] \[\therefore \] Amplitude of SHM = 0.25m
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