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Homework Help: Harmonic motion

  1. Mar 31, 2005 #1
    I have an object number 1 (M1) and second object number 2 (M2).
    Mass of number 1 is known and number 2 is unknown.
    And M1 << M2. Both massess are attached to a spring with spring constant k, where k is known. Also the period T is known. How do I solve mass number 2?
    The system:

    No gravitation, mass of the spring is negligible and so is friction in the system.
  2. jcsd
  3. Mar 31, 2005 #2
    Perhaps you can draw the relationship from this:

    [tex] \omega = \sqrt{\frac{k}{m}} [/tex]
  4. Apr 1, 2005 #3
    What is w after all? I have not encountered it before. Has it something to do with frequency? Because
    [tex] \omega = \frac{2 \pi}{T} [/tex]
  5. Apr 1, 2005 #4


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    The "w" WAS [tex]\omega[/tex], the frequency.
  6. Apr 1, 2005 #5
    What do you call period T? Because the force of the swing on both ends will be the same, the two bodies with different masses will oscillate at different frequencies.
  7. Apr 1, 2005 #6
    Period T is the period of oscillation of M1's.
    Here is the correct answer (needs explaining though):

    "The system oscillates like an object with mass:
    [tex] m_0 = \frac{M_1 M_2}{M_1 + M_2} [/tex]
    Attached to an end of a spring (the other end is fixed)."

    Where does that come from? With this we can solve M2.
  8. Apr 1, 2005 #7
    ramolarri said:

    "....the two bodies with different masses will oscillate at different frequencies."

    Not true, because if they did, the center of mass would be oscillating, which would require an outside force, and there isn't one. Both masses oscillate at the same frequency, and T is the period of the oscillation.

    King, w (omega) is called the angular frequency, and yes, it's related to frequncy (which is 1/T): w = 2pi x frequency. It turns out that angular frequency is usually a little easier to work with than 1/T.

    To understand where that m1m2/m1+m2 comes from, imagine the system oscillating/ Now is there any point on the spring that isn't moving? Clearly it's not either of the ends of the spring, because that's where the masses are, and they're moving. But is there some place along the length of the spring that never moves?
  9. Apr 1, 2005 #8
    What you're talking about is a two body oscillator. But before that you need to know the basics of simple harmonic motion. Things like frequency, amplitude and time period must be well understood to be able to solve the problem nonmechanically.

    As you already have the answer, you need to understand two things:

    1. The expression for [itex]m_{0}[/itex] is symmetric in [itex]M_1[/itex] and [itex]M_2[/itex].
    2. The system can now be replaced by a SINGLE block of mass [itex]m_0[/itex] and a SINGLE spring (of the same spring constant you started out with of course).

    [itex]m_0[/itex] is called the reduced mass of the twobody oscillator. This system has several applications not only when confined to the linear restoring force given by Hooke's Law, but in the general Two Body Problem in Mechanics (a closed form solution to which is possible). In particular, reduced mass considerations feature in gravitation, the generalized Bohr Theory for the hydrogen atom and of course, in quantum mechanics. The basic idea is simple in this case because you are essentially replacing a complex oscillator with a single equivalent spring+mass system which behaves (responds) in exactly the same way. Once you have the solution to the equivalent system, it is fairly easy to find dynamic variables for the individual problems.

    But I do think you should solve this problem only when you have understood the basics of oscillatory/periodic motion well. That way you'll appreciate what has been said here.

    **The following link might be useful (though not immediately relevant unless you can adapt it to the present problem by making a few intelligent changes)


    Hope that helps...

  10. Apr 1, 2005 #9

    I understand simple harmonic motion. For me the real problem was to find the expression for [itex]m_{0}[/itex]. The rest is easy even for me. I think a system with two masses is far from simple... :smile:

    In school we've never used this omega, what has the 2pi to do with movement only in y-axis anyway?
    Thanks for long answers, you can post more if you want. All stuff related to harmonic motion of many springs/masses is welcome. In the meantime, I'll start thinking over the last two posts.


    Okay, "reduced mass" seems to be the key word. I'll start to become familiar with it..
    Last edited: Apr 1, 2005
  11. Apr 1, 2005 #10
    King, the [itex]2\pi[/itex] appears virtually everywhere...I'll try and build up a bit:

    The differential equation (de) of motion for a mass executing SHM is of the form

    [itex]\ddot{x} + \omega^2x = 0[/itex]

    where [itex]\omega = \sqrt{\frac{K}{M}}[/itex]

    From the point of view of mathematics this merely a substitution intended to make the solution of the de look nice :smile:. The solution is (as you may verify),

    [itex]x = A\cos(\omega t + \phi)[/itex]

    If you forget about [itex]\phi[/itex] for now and consider the argument of the cos term, it involves the same omega we defined above. This is called the angular frequency of the harmonic motion. It is related to frequency f by

    [itex]\omega = 2\pi f[/itex]

    The [itex]2\pi[/itex] term appears even if motion is along a straight line. Its because of trigonometry: the cosine function is periodic with period [itex]2\pi[/itex]. In the solution I have written for x, if I replace [itex]\omega t[/itex] by [itex]\omega t + 2\pi[/itex] I should get the same value for x. This means that because of the identity,

    [itex]\omega t + 2\pi = \omega(t + \frac{2\pi}{\omega})[/itex]

    the time period of the oscillation is [itex]T = \frac{2\pi}{\omega}[/itex]. This is the time taken to execute one complete cycle of the oscillation (i.e. to come back to the same phase as one started out with--this too takes some explaining if you are unfamilar with cosine graphs so lemme know if thats the case). We also know that the number of cycles per second is the frequency, given by

    [tex]f = \frac{1}{T}[/tex]

    so this easily gives [itex]f = \frac{2\pi}{\omega}[/itex] and hence [itex]\omega = 2\pi f[/itex].

    Hope that helps...

    Last edited: Apr 1, 2005
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