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Homework Help: The period of the spring without ignoring its mass

  1. Feb 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A block of mass M is connected to a spring of mass m and oscillates in simple harmonic motion on a horizontal, frictionless track. The force constant of the spring is k and the equilibrium length is l. Assume that all portions of the spring oscillate in phase and that the velocity of a segment dx is proportional to the distance x from the fixed end; that is, vx = (x/l)v. Also, note that the mass of a segment of the spring is dm = (m/l)dx.

    (a) Find the kinetic energy of the system when the block has a speed v.
    (b) Find the period of oscillation.

    2. Relevant equations
    [tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]

    3. The attempt at a solution
    I've solved (a).

    For (b), I think I sill can, by the net force of the M block, get the SHM period.

    No matter how much mass the spring has, the net force of the M block is still :

    So, I think the period is still
    [tex]T=2\pi \sqrt{\frac{M}{k}}[/tex]

    But the answer is
    [tex]T=2\pi \sqrt{\frac{M+\frac{1}{3}m}{k}}[/tex]
    I think it's a little bit plausible because the mass in this equation is the effective mass.
    However, I really want to know what happened to the net force of M block.
    Can I still use Hooke's law? If I can't use it, then what's the meaning of spring constant k now?
    Can I solve the period without using effective mass? I want to use Hooke's law and Newton's law to solve the
    2nd order differential equation and get the answer.

    Thank you very much!
  2. jcsd
  3. Feb 19, 2015 #2


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    net Force -ks is applied to the system from the attachment point (wall), but there is more inertia than just the block M.
    You found out yourself by integration that the spring's inertia is m/3 (due to not moving as far as M).
    spring stiffness k still relates the static Force to the stretch, but of course its weight will stretch it some if it is suspended from ceiling.
    The Force BY the block M applied to the spring cannot be opposite the Force by the wall, because the spring's mass accelerates (well, some of it does)
  4. Feb 19, 2015 #3


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    To put what lightgrav said a little differently, the tension will not be uniform along the spring.
    I suggest using part (a) as a hint. They got you to find the KE. What might you consider next?
  5. Feb 19, 2015 #4
    Thanks, lightgrav and haruspex!
    I think the key is
    [tex]F_{net}=(-kx)+(+F_{pull})=ma=m\times 0=0[/tex]
    This equation still holds because the system is static!
    If I really want to write down the net force equation about block M while it's moving, then perhaps I need to use the force done by the adjacent small segment of the spring on the block M. But it's too difficult to write.. I'm not sure how to write it down because the description of effective spring constant and the displacement from the right end relaxed position is very confusing. Anyway, thanks a lot!!
  6. Feb 19, 2015 #5


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    It is? I thought you were trying to show it followed SHM. That's hardly static.
    Did you think about my KE hint?
  7. Feb 19, 2015 #6
    Yes, I guess I know what you meant.
    1. We can derive the effective mass from KE then plug in the period formula.
    2. We can derive the net force of block M by taking the derivative of KE with respect to x. Is this what you meant?
  8. Feb 19, 2015 #7


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    Not quite. I was thinking of taking the time derivative of PE+KE=constant.
  9. Feb 19, 2015 #8
    Wow, I find it works!!

    [tex]U_{s}=\lim_{n\rightarrow \infty }\sum_{i=1}^{n}(\frac{1}{2}k_{i}x_{i}^{2})[/tex]
    [tex]kl=\lim_{n\rightarrow \infty }k_{i}(\frac{l}{n})\: \therefore \: k_{i}=\lim_{n\rightarrow \infty }nk[/tex]
    Assume the total displacement is x, then
    [tex]x_{i}=\lim_{n\rightarrow \infty }\frac{x}{n}[/tex]
    [tex]\therefore \: U_{s}=\lim_{n\rightarrow \infty }\sum_{i=1}^{n}\frac{1}{2}(nk)(\frac{x}{n})^{2}=\lim_{n\rightarrow \infty }\sum_{i=1}^{n}\frac{1}{2n}kx^{2}=\lim_{n\rightarrow \infty }n\times \frac{1}{2n}kx^{2}=\frac{1}{2}kx^{2}[/tex]
    [tex]\therefore U_{s}+E_{k,total}=\frac{1}{2}kx^{2}+\frac{1}{2}(M+\frac{1}{3}m)v^{2}=constant[/tex]
    [tex]\therefore \: \frac{d(U_{s}+E_{k,total})}{dt}=kx\frac{dx}{dt}+(M+\frac{1}{3}m)v\frac{dv}{dt}=0[/tex]
    [tex]\therefore \: -kx=(M+\frac{1}{3}m)\frac{d^{2}x}{dt^{2}}[/tex]
    [tex]\therefore T=2\pi \sqrt{\frac{(M+\frac{1}{3}m)}{k}}[/tex]

    It seems that I can't get that Newton's 2nd law equation without taking the time derivative of PE+KE=constant.
    Furthermore, I doubt that -kx is a kind of force. It seems to me that it's merely a mathematical equation without physical meaning such as force (as the cause of motion).
    Is -kx a kind of force? Can I draw the free body diagram? What kind of system has the mass M+m/3? Maybe the only system is just "equivalent system".
    Thanks a lot, haruspex!
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