A 50.0 cm long organ pipe is filled with air at 10.0°C and is open at both ends. The speed of sound in air at 0°C is 331 m/s. Another 50.0 cm long organ pipe is filled with air at 15.0°C and is open at both ends. What is the beat frequency between the fundamental modes of the two organ pipes?

Question

Quizplus · Accepted Answer

The speed of sound in air increases with temperature, following the formula $v = v_0 \sqrt{\frac{T + 273}{273}}$, where $v_0$ is the speed of sound at 0°C (331 m/s), and $T$ is the temperature in °C. For the first organ pipe at 10.0°C, the speed of sound is $v_1 = 331 \sqrt{\frac{10 + 273}{273}}$, and for the second organ pipe at 15.0°C, it's $v_2 = 331 \sqrt{\frac{15 + 273}{273}}$. The fundamental frequency ($f$) of an open pipe is given by $f = \frac{v}{2L}$, where $L$ is the length of the pipe. Thus, the fundamental frequencies for the two pipes are $f_1 = \frac{v_1}{2 \times 0.5}$ and $f_2 = \frac{v_2}{2 \times 0.5}$, respectively. The beat frequency is the absolute difference between these two frequencies, $|f_1 - f_2|$. Calculating these values gives a beat frequency of approximately 2.96 Hz.

A 500 Cm Long Organ Pipe Is Filled with Air at at 10.0°C