Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 11 - Oscillations and Waves - Problems - Page 325: 56

Answer

Three standing wave patterns.

Work Step by Step

The tension in the string is the weight of the mass on the end. $$F_T=mg$$ We may find the speed of waves on the string. $$v=\sqrt{\frac{F_T}{\mu}}=\sqrt{\frac{mg}{\mu}}$$ Next, we find the wavelength of the waves, using the given frequency of 60.0 Hz. $$\lambda=\frac{v}{f}=\frac{1}{f}\sqrt{\frac{mg}{\mu}}$$ $$\lambda=\frac{v}{f}=\frac{1}{60.0Hz}\sqrt{\frac{(0.080kg)(9.80\;m/s^2)}{3.5\times10^{-4}\;kg/m}}=0.7888\;m$$ For there to be nodes at both ends and to form a standing wave, the length of the string must be an integer multiple of half a wavelength. The possible lengths are $\lambda/2$, $\lambda$, $3\lambda/2$, $2\lambda$... This means the possibilities for lengths are 0.39m, 0.79m, 1.18m, 1.58m, etc. If the length varies between 0.10m and 1.5m, only the first 3 solutions are valid.
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