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SHM question

  1. Aug 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Two masses m1 and m2 are joined by a light spiral spring. Each body oscillates along the axis of the spring, which obeys Hooke's law in both extension and compression. The bodies move in opposite directions and the centre of mass of the system is stationary. Explain why the periods of the oscillations of both bodies are the same

    2. Relevant equations

    3. The attempt at a solution
    Since the two masses are attached to the same spring, they should exprience the same amount of restoring force F.

    Hence, F=m1a1=m2a2=kx
    Since m1 and m2 are not the same, the value of k/m1 and k/m2 should also not be the same.
    T = 2pi(k/m)1/2
    Following this reasoning, how can the periods of oscillations of both body be the same? I don't understand.
  2. jcsd
  3. Aug 23, 2010 #2
    no actually ur working is wrong. considering centre of mass as a reference frame, the length from the cm to m1 will be m2*x/(m1+m2) and to m2 will be m1*x/(m1+m2). now spring constant k1 for m1 will be (m1+m2)/m2*k and k2 will be (m1+m2)/m1*k. so the (omega)^2 for both will be ((m1+m2)/m1m2). hence the time period for both will be 2pi/(omega) which is same for both.
  4. Aug 23, 2010 #3
    Why are you considering the distance from centre of mass but not the total extension of the spring? I thought it is the total extension that provides the restoring force?
  5. Aug 23, 2010 #4
    first thing I would like to ask u is abt which point these two masses will oscillate??? u need to consider a fixed point for this and hence is the com.
  6. Aug 23, 2010 #5
    Sorry, i did not read your previous thread carefully. Now I understood what you mean. But i am still a bit uncomfortable with the fact that the same spring will have different spring constant.
  7. Aug 23, 2010 #6
    it is not that the "same spring will have different spring constant", but different length of the same spring will have different spring constant. Suppose u have one string of 1 m and k=1. so if u break it like 2/3 from say right. Then the spring constant of the right spring will be 3k/2 will for the left part will be 3k.
    mathematically; suppose F/x=k
    then F/(2x/3)=3/2 k
    I hope u got it now =p
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