A string with a mass per length of $2.00 \mathrm{~g} / \mathrm{m}$ is stretched with a force of $150 \mathrm{~N}$ between two points that are $0.500 \mathrm{~m}$ apart. The fundamental frequency of the stretched string is in tune with the fundamental frequency 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) $50.2 \mathrm{~cm}$
B) $60.4 \mathrm{~cm}$
C) $57.5 \mathrm{~cm}$
D) $65.3 \mathrm{~cm}$
E) $55.2 \mathrm{~cm}$

Question

Quizplus · Accepted Answer

The fundamental frequency of a string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length of the string, $T$ is the tension in the string, and $\mu$ is the mass per unit length of the string. For the given string, $L = 0.500 \, \text{m}$, $T = 150 \, \text{N}$, and $\mu = 2.00 \, \text{g/m} = 0.00200 \, \text{kg/m}$. Plugging these values into the formula gives the fundamental frequency of the string.The fundamental frequency of an organ pipe open at both ends is given by $f = \frac{v}{2L}$, where $v$ is the speed of sound in air and $L$ is the length of the organ pipe. Given that the fundamental frequency of the organ pipe is in tune with that of the string, we can equate the two expressions for frequency and solve for $L$, the length of the organ pipe, using the given speed of sound in air, $v = 331 \, \text{m/s}$.Solving these equations with the given values leads to the length of the organ pipe being approximately $60.4 \, \text{cm}$, which corresponds to choice B.

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 $2.00 \mathrm{~g} / \mathrm{m}$