Question

A string of length L = 1.2 m is attached at one end to a wave oscillator, which is vibrating at a frequency f = 80 Hz. The other end of the string is attached to a mass hanging over a pulley as shown in the diagram below. When a particular hanging mass is suspended from the string, a standing wave with two segments is formed. When the weight is reduced by 2.2 kg, a standing wave with five segments is formed. What is the linear density of the string?

Answer 1

given

f = 80 Hz ,

L = 1.2 m ,

weight = 2.2 kg

say m is the linear density

now using the values

f = ( 2/ 2 L ) ( M g / m )^1/2 --- 1

= ( 5 / 2 L ) ( ( M - 2.2 ) g / m )

2 M^1/2 = 5 ( M - 2.2 )^1/2

4 M = 25 ( M - 2.2 )

M / ( M - 2.2 ) = 25 / 4

M = 2.61 kg

substituting in equation 1

80 = ( 2/ 2 X 1.2 ) ( 2.61 X 9.8 / m )^1/2

( 2.61 X 9.8 / m )^1/2 = 96.38

m = ( 2.61 X 9.8 / 96.38² )

m = 2.75 X 10^-3 kg / m