Longitudinal waves in a clamped metal rod

[jemerlia]

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?

 

[Redbelly98]

The rod length is 1.20 m, not 1.60 m.

 

[tomcue]

As a fellow scientist, I would like to provide some feedback on your solution. First, I would like to commend you for attempting to solve this problem using the correct equations. However, I believe there are some errors in your calculations.

For part (a), you correctly identified that the first overtone is the third harmonic. However, the equation you used to calculate the wavelength is incorrect. The correct equation for the wavelength of a standing wave in a clamped rod is w = (2L)/n, where L is the length of the rod and n is the harmonic number. Therefore, for the first overtone (n=3), the wavelength would be w = (2 x 1.20m)/3 = 0.8m. Using this wavelength, we can calculate the frequency using the equation v = f x w, which gives us f = 3600 m/s / 0.8m = 4500 Hz.

For part (d), you made a similar mistake in your calculation of the wavelength. The correct equation for a rod clamped in the middle would be w = L/n, where L is the length of the rod and n is the harmonic number. For the first overtone (n=3), the wavelength would be w = 1.20m/3 = 0.4m. Using this wavelength, we can calculate the frequency using the same equation, v = f x w, which gives us f = 3600 m/s / 0.4m = 9000 Hz.

In summary, it seems that you have used the wrong equations for calculating the wavelength in both parts of the problem. By using the correct equations, we get the correct answers of 4500 Hz and 9000 Hz for parts (a) and (d), respectively. I hope this helps clarify your solution. Keep up the good work!