Two taut strings of identical mass per unit length are stretched with the same tension with their ends fixed,but one string is 0.330 cm longer than the other.Waves on these strings propagate at 34.0 m/s.The fundamental frequency of the shorter string is 258 Hz.What is the beat frequency when each string is vibrating at its fundamental frequency?
A)12.3 Hz
B)9.00 Hz
C)12.0 Hz
D)12.7 Hz
E)11.3 Hz

Question

Quizplus · Accepted Answer

The beat frequency is given by the difference in frequencies of the two strings. The frequency of a string is given by:$$f = \frac{v}{2L}\sqrt{\frac{T}{\mu}}$$where v is the wave speed, L is the length of the string, T is the tension, and $\mu$ is the mass per unit length.Since the two strings have the same mass per unit length and tension, the only difference in their frequencies is due to the difference in their lengths. The shorter string has a frequency of 258 Hz, so the longer string has a frequency of:$$f_2 = f_1\frac{L_1}{L_2} = 258 	ext{ Hz}\frac{L_1}{L_1 + 0.0033 	ext{ m}} = 258 	ext{ Hz}\frac{1}{1.0033} = 257.1 	ext{ Hz}$$The beat frequency is then:$$f_b = |f_1 - f_2| = |258 	ext{ Hz} - 257.1 	ext{ Hz}| = 0.9 	ext{ Hz}$$

Two Taut Strings of Identical Mass Per Unit Length Are