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Manipal Medical Manipal Medical Solved Paper-2011

  • question_answer
    A particle is executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its displacement is 0.02 and its acceleration is\[0.5\text{ }m/{{s}^{2}}\]. The maximum velocity of the particle is (in m/s)

    A)  0.01              

    B)  0.05

    C)  0.5               

    D)  0.25

    Correct Answer: C

    Solution :

     Maximum acceleration \[\alpha ={{\omega }^{2}}y\] \[\therefore \] \[0.5={{\omega }^{2}}\times 0.02\] Or \[{{\omega }^{2}}=\frac{0.5}{0.02}=25\] So, \[\omega =5\] Now, maximum velocity is \[v=a\omega =0.1\times 5=0.5\,m/s\]


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