A 1.7-m string is held fixed at both ends.When driven by a 165-Hz source,the string vibrates in 2 distinct segments.What is the natural fundamental frequency of the string?

Question

Quizplus · Accepted Answer

If a string vibrates in 2 distinct segments, it means that there is one node (point of zero amplitude) in the middle of the string. This indicates that there are two wavelengths (or one wavelength and one-half wavelength) in the string.

Let's consider the case where there is one wavelength in the string. The length of the string is 1.7 m, which means that the wavelength is twice the length or 3.4 m. Using the formula v = fλ (where v is the velocity of the wave and f is the frequency), we can solve for the velocity of the wave:
v = fλ
v = f(2L)
v = f(2 x 1.7)
v = 3.4f

We know that the source has a frequency of 165 Hz, so the velocity of the wave is:
v = 3.4 x 165
v = 561 m/s

Now we can calculate the fundamental frequency of the string (i.e. the frequency at which the string will vibrate with one full wavelength):
f = v/λ
f = v/(2L)
f = 561/3.4
f ≈ 165 Hz

However, this is the frequency of the source, and not the fundamental frequency of the string. The fundamental frequency is given by:
f1 = v/(2L)
f1 = 561/(2 x 1.7)
f1 ≈ 83 Hz

Therefore, the answer is A) 83 Hz.

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