A 30.0 cm long organ pipe is filled with air and is open at one end and closed at the other. The speed of sound in air at 0°C is 331 m/s. What is the frequency of the second mode of vibration?
A) 900 Hz
B) 828 Hz
C) 750 Hz
D) 721 Hz
E) 650 Hz

Question

Quizplus · Accepted Answer

For a pipe open at one end and closed at the other, the harmonics are odd multiples of the fundamental frequency. The fundamental frequency (first harmonic) corresponds to a quarter wavelength fitting in the pipe. The second mode of vibration for such a pipe is the third harmonic, which corresponds to three quarters of a wavelength fitting in the pipe. The wavelength $\lambda_1$ for the fundamental frequency can be found using the length of the pipe $L = \frac{1}{4}\lambda_1$, so $\lambda_1 = 4L$. For the third harmonic, the wavelength $\lambda_3 = \frac{4L}{3}$.Given that $L = 30.0 \, \text{cm} = 0.30 \, \text{m}$, and the speed of sound $v = 331 \, \text{m/s}$, the frequency of the third harmonic (second mode of vibration) can be found using $f = \frac{v}{\lambda}$.$$f_3 = \frac{v}{\lambda_3} = \frac{v}{\frac{4L}{3}} = \frac{3v}{4L}$$Substituting the given values:$$f_3 = \frac{3 \times 331}{4 \times 0.30} = \frac{993}{1.2} = 827.5 \, \text{Hz}$$Rounding to the nearest whole number gives $828 \, \text{Hz}$, which corresponds to choice B.

A 300 Cm Long Organ Pipe Is Filled with Air and and Is