A violin with string length 32 cm and string density $1.5 \mathrm {~g} / \mathrm { cm }$ resonates in its fundamental with the first overtone of a 2.0-m organ pipe with one end closed and one end open.What is the tension in the string if the speed of sound in air is 344 m/s?
A)1000 N
B)110 N
C)450 N
D)4100 N
E)56 N

Question

Quizplus · Accepted Answer

The fundamental frequency of the organ pipe (first overtone for a pipe with one end closed) is given by $f = \frac{v}{4L}$, where $v$ is the speed of sound in air and $L$ is the length of the pipe. For the organ pipe, $L = 2.0 \, \text{m}$, so $f = \frac{344 \, \text{m/s}}{4 \times 2.0 \, \text{m}} = 43 \, \text{Hz}$.The frequency of the vibrating string is also 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 linear mass density of the string. Given that the string length $L = 0.32 \, \text{m}$ and the linear mass density $\mu = 1.5 \, \text{g/cm} = 0.015 \, \text{kg/m}$, we can rearrange the formula to solve for $T$:$$43 = \frac{1}{2 \times 0.32} \sqrt{\frac{T}{0.015}}$$Squaring both sides and solving for $T$ gives:$$T = 0.015 \times (2 \times 0.32 \times 43)^2 = 1000 \, \text{N}$$Therefore, the tension in the string is 1000 N.

A Violin with String Length 32 Cm and String Density 1.5 g/cm1.5 \mathrm {~g} / \mathrm { cm }1.5 g/cm

A Violin with String Length 32 Cm and String Density $1.5 \mathrm {~g} / \mathrm { cm }$