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MGIMS WARDHA MGIMS WARDHA Solved Paper-2005

  • question_answer
    A particle executing simple harmonic motion has a time peiod of 4 sec. After how much                internal of time from \[t=0\]will its displacement be half of its amplitude :

    A) \[\frac{1}{3}\]sec                            

    B) \[\frac{1}{2}\]sec

    C)  \[\frac{2}{3}\] sec                          

    D) \[\frac{1}{6}\] sec

    Correct Answer: A

    Solution :

    Time period \[T=4\text{ }\sec \] From equation of SHM \[y=a\sin \frac{2\pi }{T}.t\] Here \[y=\frac{a}{2}\] \[\therefore \]  \[\frac{a}{2}=a\sin \frac{2\pi t}{4}\] \[\frac{1}{2}=\sin \frac{\pi t}{2}\Rightarrow \sin \frac{\pi }{6}=\sin \frac{\pi t}{2}\] \[\therefore \]  \[\frac{\pi }{6}=\frac{\pi t}{2}\Rightarrow \tau =\frac{1}{3}\sec \]


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