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Homework Help: SHM Problem

  1. Aug 23, 2010 #1
    1. The problem statement, all variables and given/known data

    In a frictionless plane, a block of mass [tex]M[/tex] is attached to a spring initially relaxed of constant [tex]k[/tex] which is attached to a wall. A piece of gum of mass [tex]m[/tex] is then horizontally thrown in direction to the block with velocity [tex]v_1[/tex] hitting it at time [tex]t = 0[/tex] in a perfectly inelastic collision. Find the expression for the displacement of the system at time [tex]t > 0[/tex]. The mass of the spring is not to be taken into consideration.

    2. Relevant equations

    1. SHM equation for the displacement of the system:

    [tex]x(t) = Asin(wt) + Bcos(wt) [/tex]

    2. SHM equation for the velocity of the system:

    [tex]\dot{x}(t) = Awcos(wt) - Bwsin(wt) [/tex]


    [tex]w = \sqrt{\frac{k}{(M+m)}} [/tex]

    3. Equation of Conservation of Momentum for the system:

    [tex]p_{1} = p_{2},\ mv_1 = (m+M)v_2 [/tex]

    3. The attempt at a solution

    First of all, from the third equation we know that

    [tex]v_2 = \frac{mv_1}{(m+M)} [/tex]

    The initial conditions for the system are easily found by using [tex]t = 0[/tex] at equations 1 and 2, so

    [tex]x(0) = Asin(w \times 0) + Bcos(w \times 0) = B = 0[/tex]

    [tex]\dot{x}(0) = v_2 = Awcos(w \times 0) - Bwsin(w \times 0) = Aw[/tex]


    [tex]A = \frac{v_2}{w} [/tex]

    [tex]A = \frac{mv_1}{(m+M)} \sqrt{\frac{m}{k}} [/tex]

    So the equation for the displacement of the system at any given time(t) shall be

    [tex]x(t) = \frac{mv_1}{(m+M)} \sqrt{\frac{m}{k}}sin(\sqrt{\frac{k}{m}}t) + Bcos(\sqrt{\frac{k}{m}}t) [/tex]

    OK, so I know that 'A' shall be constant for any given time as it depends on the initial velocity of the system (which depends on the constant velocity of the piece of gum and the mass of the system) and on 'w' and both are constant. Now, how do I find 'B' for any time [tex] t > 0[/tex]? I know that at time [tex]t = 0[/tex] 'B' is zero but it won't necessarily be zero at [tex]t > 0[/tex] or am I wrong?

    Thanks in advance.
    Last edited: Aug 23, 2010
  2. jcsd
  3. Aug 23, 2010 #2


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    You treat the problem as a perfectly inelastic collision although the problem says "perfectly elastic". Is it a typo? Assuming that it is and the masses stick together after the collision, your analysis is almost correct. Constant B must be set equal to zero, otherwise x(0) is not equal to zero. In other words, you did not use the second initial condition on the position to find the second arbitrary constant. You also need to fix the frequency of oscillations ω. What is the mass at the end of the spring after the gum sticks to the block?

    *** On Edit ***
    It is a good assumption that the collision is perfectly inelastic. Gum tends to be sticky and a perfectly elastic collision with it is highly unlikely.
    Last edited: Aug 23, 2010
  4. Aug 23, 2010 #3
    first of all, I think it should be an inelastic collision. If it is so,
    Then as u have done above u can find the v of the system. This will be vmax. Hence equate the total kinetic energy of the system with the with the total potential energy 0.5 *k *x^2. The x u will get now is the ampitude. W u found out is correct. Now use the equation x(t)=Asin(wt) to get the displacement. I dont now why u have said it should be Asin(wt)+Bcos(wt)
  5. Aug 23, 2010 #4


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    It is a perfectly elastic collision, so your equation for the velocity after collision

    p_{1} = p_{2},\ mv_1 = (m+M)v_2

    is not valid.

  6. Aug 23, 2010 #5
    Yes, it is indeed a perfectly inelastic collision I mistyped it but already corrected it.

    So 'B' is in fact zero?

    I know that 'B' is zero at [tex]t = 0[/tex], I've shown it in the work done, but why does it equal zero at also [tex]t > 0[/tex], what proves this? This is what I'm not getting.

    That way I don't have to deal with the phase angle in [tex]x(t) = Amplitude \times cos(wt + \phi) [/tex], it is just another way of working it out.

    It is a perfectly inelastic collision, I've already corrected it.
  7. Aug 23, 2010 #6
    from what I know we use Asin(wt) when u start counting ur time from the equilibrium position while Acos(wt) for time starting from the amplitude of the SHM. U dont need to worry about phase constant here since the system is starting from its mean(equilibrium position).
  8. Aug 23, 2010 #7


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    Yes, constants A and B are arbitrary. The general solution is

    x(t) = A sin(ωt) + B cos(ωt)

    They are what is necessary to fit the initial conditions namely the specific values of x(t=0) and v(t=0) suitable to a specific problem. Two initial conditions, two constants to specify. Whether you use A and B or an amplitude and a phase angle, you still need to specify two constants for the two initial conditions.
  9. Aug 23, 2010 #8
    You use your initial conditions to find the coefficients A and B. You have the correct value for A, and we see that B is 0. Hence, this X(T)=Asin(wt) satisfies the initial condition X(0)=0. We know X(0)=0 because the problem says the spring is relaxed.

    Look at it this way... If it had said the Spring had been displaced .1m beyond its equilibrium point at t=0, then B would have been non-zero: X(0)= B= 0.1.

    Furthermore, B is a constant, its not just 0 at t=0, and non-zero for other values of t. That would imply B is a function of t, and thats obviously not true.
  10. Aug 23, 2010 #9
    Yes, my whole confusion was that I was thinking of 'B' as a VARIABLE rather than a CONSTANT. As it is a CONSTANT, by definition it will hold the same value for every time (t), so if you find the value for 'B' at [tex]t = 0[/tex] it will remain this same value for every [tex]t > 0[/tex] heh.

    Thank you very much for your time and patience kuruman, Swap, ehild and aq1q.

    Problem solved. :smile:
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