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 string is given by:$$f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$where L is the string length, T is the tension, and $\mu$ is the string density.The first overtone of the organ pipe is given by:$$f_2 = \frac{3}{4L} v$$where L is the pipe length and v is the speed of sound in air.Since the string and the organ pipe resonate in their fundamental and first overtone, respectively, we have:$$f_1 = f_2$$$$\frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{3}{4L} v$$Solving for T, we get:$$T = \frac{9}{16} \mu v^2$$Substituting the given values, we get:$$T = \frac{9}{16} (1.5 	imes 10^{-3} \mathrm{~kg/cm}) (344 \mathrm{~m/s})^2 = 1000 \mathrm{~N}$$

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 $1.5 \mathrm {~g} / \mathrm { cm }$