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Energy conservation vs momentum conservation in SHM

  1. Feb 3, 2012 #1
    1. The problem statement, all variables and given/known data
    a mass M , attached to a horizontal spring executes SHM(simple harmonic motion) with amplitude A1 . when the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude A2 . the ratio A1/A2 is ...?

    2. Relevant equations
    taking angular frequency = ω

    3. The attempt at a solution

    first taking two mass and spring as the system, since there is no external force momentum remain conserved , applying (M)(ω1)(A1) = (M+m)(ω2)(A2) , i get correct answer but while applying energy conservation equations, i get an incorrect answer!
    will energy not remain conserved in such process?

    plz help!
  2. jcsd
  3. Feb 3, 2012 #2


    User Avatar
    Homework Helper
    Gold Member

    Hi nikhilarora,

    You might have done something wrong. Show your work.

  4. Feb 3, 2012 #3
    using k=spring constant
    applying momentum conservation :

    M*(ω1)*A1 = (m+M)*(ω2)*A2
    =>M*(k/M)1/2*A1 = (m+M)*(k/m+M)1/2*A2

    => A1/A2 = (M+m/M)1/2

    applying energy conservation :

    1/2 * M*(ω1)2 * A12 = 1/2 * (m+M) * (ω2)2 * A22
    => M*(k/M)*A12 = (m+M)*(k/m+M)*A22
    => A12=A22
    => A1=A2

    where have i done wrong ??
  5. Feb 3, 2012 #4


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    Homework Helper
    Gold Member

    I see. The energy is not conserved when you put the small mass on the vibrating one. It is like an inelastic collision: the masses move together. So the bigger mass has to accelerate up the smaller one, and that happens with the assistance of friction or some other force which does work.

  6. Feb 3, 2012 #5
    thanks a lot !!
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