A string with a mass per length of 2.00 g/m is stretched with a force of 150 N between two points that are 0.500 m apart. The fundamental frequency of the stretched string is in tune with the fundamental frequency of an organ pipe filled with air at 20°C and open at both ends. The speed of sound in air at 0°C is 331 m/s. What is the length of the organ pipe?
A) 70.5 cm
B) 62.6 cm
C) 59.3 cm
D) 53.8 cm
E) 50.2 cm

Question

Quizplus · Accepted Answer

The fundamental frequency of the string can be found using the formula $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, 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. Substituting the given values, $L = 0.500 \, \text{m}$, $T = 150 \, \text{N}$, and $\mu = 2.00 \, \text{g/m} = 0.002 \, \text{kg/m}$, we find the frequency of the string. The speed of sound in air at 20°C can be found using the formula $v = v_0 + 0.6T$, where $v_0 = 331 \, \text{m/s}$ is the speed of sound at 0°C and $T$ is the temperature in °C. At 20°C, the speed of sound is $v = 331 + 0.6(20) = 343 \, \text{m/s}$. The fundamental frequency of an open pipe is given by $f = \frac{v}{2L}$, where $v$ is the speed of sound in air and $L$ is the length of the pipe. Equating the frequency of the string to that of the organ pipe and solving for $L$ gives the length of the organ pipe that matches the fundamental frequency of the string.

A String with a Mass Per Length of 2