- #1

- 4

- 0

M+[tex]\frac{m_{s}}{3}[/tex]

Derive an expression for [tex]m_{s}[/tex].

------------------------------------

Well first off:

U=[tex]\frac{1}{2}[/tex]k[tex]x^{2}[/tex]

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

& E = K + U ,where E is constant

So I just get the time derivative:

[M+[tex]\frac{m_{s}}{3}[/tex]]a + kv = 0 ,where

[tex]v = \omega A cos \omega t[/tex]

&

[tex]a = -\omega^{2}A sin \omega t[/tex]

Then I just solve for [tex]m_{s}[/tex]

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