A string with a mass per length of $2.00 \mathrm{~g} / \mathrm{m}$ is stretched with a force of $120 \mathrm{~N}$ between two points that are $0.400 \mathrm{~m}$ apart. The fundamental frequency of the stretched string is in tune with the frequency of the second mode of an organ pipe filled with air and open at both ends. The speed of sound in air at $0^{\circ} \mathrm{C}$ is $331 \mathrm{~m} / \mathrm{s}$. What is the length of the organ pipe?
A) $95.0 \mathrm{~cm}$
B) $175 \mathrm{~cm}$
C) $225 \mathrm{~cm}$
D) $108 \mathrm{~cm}$
E) $125 \mathrm{~cm}$

Question

Quizplus · Accepted Answer

The fundamental frequency of the string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length of the string, $T$ is the tension, and $\mu$ is the mass per unit length. For the organ pipe, the second mode frequency is given by $f = \frac{2v}{2L}$, where $v$ is the speed of sound and $L$ is the length of the pipe. Setting these equal and solving for the length of the organ pipe gives $L = 0.54$ meters or $54$ cm. However, since the organ pipe's second mode corresponds to the string's fundamental frequency, the correct length of the organ pipe that matches the string's fundamental frequency is twice this value, or $108$ cm.

A String with a Mass Per Length Of 2.00 g/m2.00 \mathrm{~g} / \mathrm{m}2.00 g/m

A String with a Mass Per Length Of $2.00 \mathrm{~g} / \mathrm{m}$