A) \[\frac{1}{3}s\]
B) \[\frac{1}{2}s\]
C) \[\frac{1}{4}s\]
D) \[2s\]
Correct Answer: A
Solution :
\[y=A\sin (\omega t+\phi )\] ..(i) At extreme position, \[y=A\] and \[t=0\] So, \[A=A\sin (0+\phi )\] \[\Rightarrow \] \[\phi =\frac{\pi }{2}\] From Eq. (i), \[y=A\sin \left( \omega t+\frac{\pi }{2} \right)=A\cos \omega t\] At half the amplitude i.e. at \[y=\frac{A}{2}\] \[\frac{A}{2}=A\cos \omega t\] \[\Rightarrow \] \[\omega t={{\cos }^{-1}}\left( \frac{1}{2} \right)\] \[\Rightarrow \] \[\frac{2\pi }{T}t=\frac{\pi }{3}\] \[\Rightarrow \] \[\frac{2}{2}t=\frac{1}{3}\] \[t=\frac{1}{3}s\]
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