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?

Explanation:

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

n =

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 =            I

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

Thus n =          II

Dividing II by I , we have

n/420 = 5/3

or n = 5/3 x 420 = 700 Hz

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2