How Do You Calculate the Mass of a Spring in Simple Harmonic Motion?

[Pee-Buddy]

Hi there, I just want to confirm my answer: A mass "M" is set oscillating on a spring of mass "m_{s}". If the total mass of the system is given by:
M+\frac{m_{s}}{3}

Derive an expression for m_{s}.

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Well first off:

U=\frac{1}{2}kx^{2}

K=\frac{1}{2}[M+\frac{m_{s}}{3}]v^{2}

& E = K + U ,where E is constant

So I just get the time derivative:


[M+\frac{m_{s}}{3}]a + kv = 0 ,where
v = \omega A cos \omega t
&
a = -\omega^{2}A sin \omega t

Then I just solve for m_{s}

I'm pretty sure it's right, but I'd just like some confirmation.

 

[member 553083]

Yes, your solution is correct. To solve for m_s, you can rearrange the equation to:m_s = 3(-[M+\frac{m_{s}}{3}]\omega^2A sin \omega t - kv)/kvThen you can plug in the values for A, ω, k, and v to get the value of m_s.