A violin with string length 32 cm and string density 1.5 g/cm resonates with the first overtone from a 2.0-m long organ pipe with one end closed and the other end open. What is the tension in the string?
A) 1000 N
B) 110 N
C) 450 N
D) 4100 N
E) 56 N

Question

Quizplus · Accepted Answer

The first overtone of a closed-open organ pipe has a frequency of 3v/4L, where v is the speed of sound and L is the length of the pipe. The frequency of the first overtone of the violin string is given by f = (1/2L) * sqrt(T/rho), where L is the length of the string, T is the tension in the string, and rho is the density of the string. Setting these two frequencies equal, we get:3v/4L = (1/2L) * sqrt(T/rho)Solving for T, we get:T = 9rhov^2/16Plugging in the given values, we get:T = 9 * 1.5 g/cm * (343 m/s)^2 / 16T = 1000 NTherefore, the tension in the string is 1000 N.

A Violin with String Length 32 Cm and String Density