Standing sound waves: A string, 0.28 m long and vibrating in its third harmonic, excites an open pipe that is 0.82 m long into its second overtone resonance. The speed of sound in air is 345 m/s. The speed of transverse waves on the string is closest to
A) 120 m/s.
B) 110 m/s.
C) 100 m/s.
D) 98 m/s.
E) 91 m/s.

Question

Quizplus · Accepted Answer

The third harmonic of a string corresponds to $n = 3$, where $n$ is the harmonic number. The wavelength $\lambda_s$ for the string can be found using the formula $\lambda = \frac{2L}{n}$, where $L$ is the length of the string. Substituting $L = 0.28$ m and $n = 3$, we get $\lambda_s = \frac{2 \times 0.28}{3} = \frac{0.56}{3} = 0.1867$ m. The frequency $f$ can be found using the speed of sound $v = 345$ m/s and the wavelength $\lambda_p$ for the pipe in its second overtone (third harmonic, $n = 3$ for an open pipe). The wavelength for the pipe is $\lambda_p = \frac{2L}{n} = \frac{2 \times 0.82}{3} = \frac{1.64}{3} = 0.5467$ m. The frequency is $f = \frac{v}{\lambda_p} = \frac{345}{0.5467} \approx 631$ Hz. The speed of transverse waves on the string is $v_s = f \times \lambda_s = 631 \times 0.1867 \approx 118$ m/s, which is closest to 120 m/s.

Standing Sound Waves: a String, 0