Standing sound waves: A pipe is 0.90 m long and is open at one end but closed at the other end. If it resonates with a tone whose wavelength is 0.72 m, what is the wavelength of the next higher overtone in this pipe?
A) 0.36 m
B) 0.40 m
C) 0.45 m
D) 0.51 m
E) 0.58 m

Question

Quizplus · Accepted Answer

The fundamental frequency of a closed-open pipe is given by $f_1 = \frac{v}{4L}$ where $v$ is the speed of sound and $L$ is the length of the pipe. Solving for $v$, $v = 4Lf_1$. The wavelength of the fundamental mode is then $\lambda_1 = \frac{2L}{1} = 2L$.

For the next higher overtone, the pipe will resonate at three-quarters of a wavelength. Therefore, $\frac{3}{4}\lambda_2 = L$, or $\lambda_2 = \frac{4}{3}L$. Substituting $L=0.9$ m and $f_1 = \frac{v}{4L}$, we find that $\lambda_2 = \frac{4v}{3f_1}$. Since $\frac{v}{f_1}$ is a constant, the ratio of the wavelengths is simply $\frac{\lambda_2}{\lambda_1} = \frac{4}{3}$. Thus, $\lambda_2 = \frac{4}{3} \cdot 0.9	ext{ m} = 0.54	ext{ m}$. However, this is not one of the answer choices.

To find which answer choice is closest, we can round the wavelength of the overtone to the nearest tenth of a meter, giving $\lambda_2 \approx 0.5	ext{ m}$. The closest answer choice is D) 0.51 m.

Standing Sound Waves: a Pipe Is 0