Standing Waves

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A string 3.2 m long and with a linear mass density of .008 kg/m is kept under tension so that traveling waves propagate at 48 m/s along the string. The ends of the string are clamped and the string vibrates in its third harmonic with an amplitude of 5.0 cm. How much energy is stored in this vibrating system at that time? If the amplitude of the standing wave diminishes to 3.0 cm in 1.0 s, what is the Q of this vibrating system?

I was thinking about starting with

Delta K = .5 U (partial(y)/partial(t))^2 delta x

but this is just a guess; could someone help me with this problem?

Thanks
 
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Answers and Replies

  • #2
kuruman
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Start with the standing wave equation$$y(x,t)=2A\sin \left(\frac{2\pi x}{\lambda} \right)\cos (\omega t)$$Here, ##\lambda=\frac{2}{3}L## and ##\omega=\frac{2\pi v}{\lambda}##.

Model the string as a series of harmonic oscillators, each of length ##dx##, mass ##dm=\rho dx## oscillating with same frequency ##\omega## and having variable amplitude ##a(x)=2A\sin \left(\frac{2\pi x}{\lambda} \right)##. At ##t=0## all the oscillators on the string are instantaneously at rest and the contribution of one of them to the potential energy is
$$dU=\frac{1}{2}(\rho dx)\omega^2 a^2= 2 \rho \omega^2 A^2 \sin^2 \left(\frac{2\pi x}{\lambda} \right)dx$$Integrate over the length of the string to get the total stored energy.
 

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