Question

A guitar string with a linear density of 2.0 g/m is stretched between supports that are 60 cm apart. The string is observed to form a standing wave with three antinodes when driven at a frequency of 420 Hz.
Part A
What is the frequency of the fifth harmonic of this string?

Answer 1

The frequency of driver is 700 Hz

Explanation:

The frequency of wave in a string is given by the relation

n = $\frac{p}{2l} \sqrt{\frac{T}{m} }$

here n is the frequency

p is the number of antinodes and l is the length of string .

T is the tension in string and m is the mass per unit length

Thus 420 =$\frac{3}{120} \sqrt{\frac{T}{2} }$            I

Now if there is 5 antinodes , the value of p = 5

Thus n = $\frac{5}{120} \sqrt{\frac{T}{2} }$          II

Dividing II by I , we have

n/420 = 5/3

or n = 5/3 x 420 = 700 Hz