2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Simple Harmonic Motion

JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Velocity of Simple Harmonic Motion

  • question_answer
    The displacement equation of a particle is \[x=3\sin 2t+4\cos 2t.\] The amplitude and maximum velocity will be respectively                                                                     [RPMT 1998]

    A)            5, 10                                         

    B)            3, 2

    C)            4, 2

    D)            3, 4

    Correct Answer: A

    Solution :

                       \[x=3\sin 2t+4\cos 2t.\]From given equation \[{{a}_{1}}=3,\ {{a}_{2}}=4,\] and \[\varphi =\frac{\pi }{2}\]            \\[a=\sqrt{a_{1}^{2}+a_{2}^{2}}\]\[=\sqrt{{{3}^{2}}+{{4}^{2}}}=5\]Þ\[{{v}_{\max }}=a\omega =5\times 2=10\]


You need to login to perform this action.
You will be redirected in 3 sec spinner