JEE Main & Advanced Physics Simple Harmonic Motion JEE PYQ-Simple Harmonic Motion

  • question_answer
    An oscillator of mass M is at rest in its equilibrium position in a potential \[V=\frac{1}{2}k{{(x-X)}^{2}}.\]A particle of mass comes from right with speed u and collides completely in elastically with and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after collisions is:\[(M=10,m=5,u=1,k=1).\] [JEE Main Online 16-4-2018]

    A) \[\frac{1}{2}\]                          

    B)  \[\frac{1}{\sqrt{3}}\]

    C)  \[\frac{2}{3}\]             

    D)      \[\sqrt{\frac{3}{5}}\]

    Correct Answer: B

    Solution :

    [b] Initial momentum of mass 'm' = mu =5
    Final momentum of system\[=(M+m)v=mu=5\]
    For second collision, mass (m=5, u = 1) coming from right strikes with system of mass 15, both momentum have opposite direction.
    \[\therefore \] net momentum = zero
    Similarly for12th collision momentum is zero.
    For 13th collision, total mass\[=10+12\times 5=70\]
    Using conservation of momentum
    \[70\times 0+5\times 1=(70+5)v'\]
    \[v'=\frac{1}{5}\]
    Total mass \[=10+13\times 5=75\]
    Finald KE of system
    \[=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times 75\times \left[ \frac{1}{15} \right]\left[ \frac{1}{15} \right]\]
    \[\frac{1}{2}k\,{{A}^{2}}=\frac{1}{2}75\times \frac{1}{15}\times \frac{1}{15}\]
    \[=\frac{1}{7}\times (1){{A}^{2}}=\frac{1}{2}75\times \frac{1}{15}\times \frac{1}{15}\]
    \[{{A}^{2}}=\frac{1}{3}\]
    \[A=\frac{1}{\sqrt{3}}\]


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