A 40.0 cm long organ pipe is filled with air and is open at both ends. The speed of sound in air at 0°C is 331 m/s. If the temperature of the air decreases by 10°C, then what is the frequency of the fundamental mode?
A) 394 Hz
B) 406 Hz
C) 462 Hz
D) 506 Hz
E) 544 Hz

Question

Quizplus · Accepted Answer

The frequency of the fundamental mode for a pipe open at both ends is given by $f = \frac{v}{2L}$, where $v$ is the speed of sound in the medium and $L$ is the length of the pipe. The speed of sound in air changes with temperature and is given by $v = v_0 \sqrt{\frac{T}{T_0}}$, where $v_0$ is the speed of sound at a reference temperature $T_0$ (in Kelvin), and $T$ is the temperature of interest (also in Kelvin). Given that $v_0 = 331$ m/s at $T_0 = 273$ K (0°C), and the temperature decreases by 10°C (to -10°C or 263 K), the new speed of sound $v$ at -10°C can be calculated as follows:$$v = 331 \sqrt{\frac{263}{273}}$$$$v \approx 324.5 \text{ m/s}$$Then, the frequency of the fundamental mode is:$$f = \frac{v}{2L} = \frac{324.5}{2 \times 0.4} = \frac{324.5}{0.8} \approx 405.625 \text{ Hz}$$Rounding to the nearest whole number gives 406 Hz, which corresponds to choice B.

A 400 Cm Long Organ Pipe Is Filled with Air and and Is