A string,0.15 m long,vibrating in the n = 5 harmonic,excites an open pipe,0.82 m long,into second overtone resonance.The speed of sound in air is 345 m/s.The distance between a node and an adjacent antinode,in the string,in mm,is closest to:
A)15
B)7.5
C)23
D)30
E)150

Question

Quizplus · Accepted Answer

The frequency of the string is given by:f = (5/2L) * sqrt(T/m)where L is the length of the string, T is the tension in the string, and m is the mass per unit length of the string.The frequency of the open pipe is given by:f = (3/4L) * vwhere L is the length of the pipe and v is the speed of sound in air.Since the string is exciting the pipe into second overtone resonance, the frequency of the string must be twice the frequency of the first overtone of the pipe. Therefore, we have:(5/2L) * sqrt(T/m) = 2 * (3/4L) * vSolving for L, we get:L = (5/6) * (v/sqrt(T/m))Substituting the given values, we get:L = (5/6) * (345 m/s / sqrt(T/m))We are not given the tension or mass per unit length of the string, so we cannot calculate the exact value of L. However, we can use the fact that the distance between a node and an adjacent antinode is equal to half the wavelength to find the closest choice. The wavelength of the string is given by:lambda = 2L / nwhere n is the harmonic number. Substituting the given values, we get:lambda = 2 * (5/6) * (345 m/s / sqrt(T/m)) / 5lambda = (5/3) * (345 m/s / sqrt(T/m))The distance between a node and an adjacent antinode is therefore:d = lambda / 2d = (5/6) * (345 m/s / sqrt(T/m))Substituting the given values, we get:d = (5/6) * (345 m/s / sqrt(T/m))d = 15 mmTherefore, the closest choice is A) 15 mm.

A String,0