Constituent traveling waves and strings and vibrations

AI Thread Summary
A 30-cm long string, clamped at one end and free at the other, vibrates in its second harmonic, leading to confusion about the wavelength of the constituent traveling waves. The equation L = n(1/4)λ was applied, initially yielding an incorrect wavelength of 60 cm when using n=2. The correct wavelength of 40 cm was found by using n=3, illustrating that only odd harmonics are allowed due to the boundary conditions. The term "constituent traveling wave" refers to the wave associated with the next harmonic number, not simply multiples of the fundamental frequency. Understanding the modes of vibration clarifies that the second harmonic in this context corresponds to the third harmonic mode.
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Homework Statement



A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wavelength of the constituent traveling waves is:
A) 10 cm
B) 30 cm
C) 40 cm
D) 60 cm
E) 120 cm

Homework Equations



L = n(\frac{1}{4}\lambda)

The Attempt at a Solution



First, what I tried was this:
.30 = 2(\frac{1}{4}\lambda)
\lambda = .60

The answer, however, was C) 40 cm.

So, when I changed 2 to 3, like this:

.30 = 3(\frac{1}{4}\lambda)
\lambda = .40

I got the correct answer.

So, does the constituent traveling wave mean the wave with the next harmonic number? Somehow, I don't think that is correct. Constituent means 'component' and such.

I don't understand the concept behind this, why do you use the harmonic number of 3?
 
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It looks like by "harmonics", they mean only the allowed vibration modes and not simply multiples of the lowest-frequency mode.

You might try drawing a sketch of the "modes" or shapes of the string for n=1, 2, and 3. See if it makes sense that the n=2 case is not valid, given the condition of 1 fixed + 1 free end.
 
Ah I see, since it's clamped at one end it can only have odd number harmonics correct? So the second harmonic they would mean n=3.
 
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