A) \[4T/3\]
B) \[3T/8\]
C) \[3T/4\]
D) \[3T/7\]
Correct Answer: B
Solution :
At phase \[3\pi /2,\,\,(\omega t+\phi )\,\,=3\pi /2\] \[{{x}_{1}}=A\,\sin \frac{3\pi }{2}\,=-A\] \[{{x}_{2}}=0\] So, at \[t=0,\,\,{{x}_{1}}=-A\,\cos \,\omega t\,\] and \[{{x}_{2}}=A\sin \omega t\] \[{{x}_{1}}={{x}_{2}}\] \[-\cos \omega t=\sin \omega t\] \[\tan \omega t=-1\] \[\omega t\,=\frac{3\pi }{4}\] \[\left( \frac{2\pi }{T} \right)\,t=\frac{3\pi }{4}\] \[t=3T/8\].You need to login to perform this action.
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