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=$$\frac{1}{2}$$k$$x^{2}$$

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.