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Change in period due to a change in mass

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data
    A mass m oscillating on a spring has a period T. Suppose the mass changes very slightly from m to m+Δm, where Δm << m. Find an expression for ΔT, the small change in the period. Your expression should involve T, m, and Δm, but NOT the spring constant (k).


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


    3. The attempt at a solution
    [tex]T+ΔT=2\pi \sqrt{\frac{m+Δm}{k}}[/tex]
    Would that be the right way to start off? I can't figure out how to get rid of the k there. I don't see how conservation of energy can help, especially considering that conservation of energy is how those equations were derived in the first place...
     
  2. jcsd
  3. Oct 24, 2012 #2

    haruspex

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    You wrote a 'relevant equation' involving T, m and k, but you haven't used it.
     
  4. Oct 24, 2012 #3
    Yes, this is a good start.
    You need to use now the fact that Δm<<m and find an approximate expression of the RHS.
    Do you know the expansion of [tex]\sqrt{1+x}[/tex] for small values of x?
     
  5. Oct 24, 2012 #4
    Yes, using the binomial approximation, I get ##\sqrt{m+Δm}=\sqrt{m} \sqrt{1+\frac{1}{2} \frac{Δm}{m}}##

    Hmm I suppose I should find T and ΔT separately next and plug those in...

    [tex]T+ΔT=2π(1+\frac{Δm}{2m})(\sqrt{m/k})=2π\sqrt{m/k} + 2π\sqrt{Δm/k}
    [/tex]

    Simplifying I get √(m/k) = 2, so T=4π.

    Solving for ΔT, I get ##ΔT=2π(1+(1/2) \frac{\Delta m}{m}) \sqrt{m/k} - T## which gets me...
    $$\Delta T = 4π + 2π \frac{\Delta m}{m} -4π= T(\frac{1}{2} \frac{\Delta m}{m})$$

    Does this look right? It seems to make sense (increasing the mass should increase period according to the starting equations). Is there any way for me to confirm this?
     
    Last edited: Oct 24, 2012
  6. Oct 24, 2012 #5
    Just realized this problem had a solution for it in the book, and that answer is right. Thanks for the help!
     
  7. Oct 24, 2012 #6
    This step is not justified by the problem and you don't need this assumption to solve the problem. 2π√(m/k) = T, where T is the initial period.
    However, the final result is OK.
     
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