A long metal rod of length L, 1/2 inch in diameter, is held between the fingers.

Question

A long metal rod of length L, 1/2 inch in diameter, is held between the fingers. It is hit on one end with a small hammer, along the axis of the rod. It resonates, emitting sound with a frequency of 6930 Hz. The amplitude of the oscillations along the bar is shown qualitatively below. What is the harmonic number of the normal mode illustrated? You have only 2 tries!

Correct, computer gets: 4

11. [1pt]
The speed of the wave is 5120 m/s. Calculate the length L of the rod.

Correct, computer gets: 1.48E+00 m

12. [1pt]
Referring to the mode above, answer T-True, F-False. E.g., if the first is T and the rest F, enter TFFF.

The rod will resonate if the fingers are 1/8 L from one end.

The wave's speed will not change if L is increased.

The frequency of that mode will decrease if L is lengthened.

The wave in the rod is transverse.

Explanation / Answer

Harmonic number is the number of half wavelengths. So there are 1/4    1/2    1/2   1/2    1/4   =     four half wavelengths. So harmonicnumber =   4

(b)   wavelength = speed / freq = 5120 / 6930 = 0.7388 meters

L = two wavelengths = 2 *0.7388 = 1.48meters

(c) F T T F

The first question is a bit ambigous so could be true.

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