RAJASTHAN PMT Rajasthan - PMT Solved Paper-2007

  • question_answer
    A particle executing simple harmonic motion has a time period of 4s. After how much interval of time from t = 0 will its displacement be half of its amplitude ?

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

    B)  \[\frac{1}{2}s\]

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

    D)         \[\frac{1}{6}s\]

    Correct Answer: A

    Solution :

    The displacement equation for \[SHM\] is                 \[y=a\sin \omega t\] where \[\omega \] is angular velocity \[\left( \omega =\frac{2\pi }{T} \right)\] and a the amplitude. Given,   \[y=\frac{a}{2},\,\,t=\frac{t}{2}\] \[\therefore \]  \[\frac{a}{2}=a\,\,\sin \frac{2\pi t}{4}\] \[\Rightarrow \]               \[\frac{1}{2}=\sin \frac{\pi t}{2}\] \[\therefore \]  \[\sin \frac{\pi }{6}=\sin \frac{\pi t}{2}\] \[\Rightarrow \]               \[\frac{\pi }{6}=\frac{\pi t}{2}\] \[\Rightarrow \]               \[t=\frac{1}{3}s\]


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