Longitudinal waves in a clamped metal rod

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SUMMARY

The discussion focuses on calculating the frequency of longitudinal waves in a clamped metal rod with a speed of sound of 3600 m/s. For the first overtone mode (third harmonic) with a rod length of 1.20 m, the correct frequency is 2250 Hz, while the user incorrectly calculated it as 1.69 kHz due to using an incorrect wavelength based on a length of 1.60 m. When clamped in the middle, the frequency for the same overtone mode with a length of 0.80 m is 4500 Hz, but the user calculated it as 3.38 kHz, indicating a misunderstanding of the harmonic relationships.

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  • Understanding of longitudinal waves and their properties
  • Knowledge of harmonic frequencies in vibrating systems
  • Familiarity with wave equations, specifically v = f × λ
  • Basic principles of sound propagation in solids
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  • Learn about wave equations and their applications in different media
  • Explore the relationship between wavelength, frequency, and speed of sound in solids
  • Investigate the effects of boundary conditions on wave behavior in rods
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Homework Statement


The speed of sound in a metal rod is 3600 m s -1. The rod is 1.20m
long and clamped at one of its ends.
(a) Determine the frequency of its vibration if longitudinal waves
are established in the rod and it is vibrating in its first overtone
mode.
(d) Determine the frequency of its vibration if clamped in the middle whilst still vibrating in the first overtone.

Homework Equations


v = f x w where w is the wavelength


The Attempt at a Solution


(a) The first overtone is the third harmonic, therefore
L = (3/4)w
= 3/4 x 1.60m
W = (4/3)L = (4/3) x 1.60m
f= 3600ms-1 / (4/3 x 1.60m) = 1.69kHz; given answer is 2250 Hz

(d) Still the third harmonic but L is now 0.8m
f = 3600ms-1 / (4/3 x 0.80m) = 3.38kHz; given answer is 4500 Hz

What have I done wrong?
 
Physics news on Phys.org
The rod length is 1.20 m, not 1.60 m.
 

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