Question

(i) For the wave on a string described in Exercise 15.11, do all the points on thebstring oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?

Answer 1

Given, the length of the string is 1.5 m and it’s mass is 3× 10 −2  kg.

i)

For the waves on the string, the length of the string is 1.5 m and its wavelength is λ=3 m. So, it is clear that the length of the string is equal to half of the wavelength of the string. For a string that is clamped at both ends, it is possible that both ends behave as nodes and there are antinodes in between.

a)

All the string particles vibrate with the same frequency f=60 Hz, except the nodes. This is because the nodes are fixed points and they do not vibrate.

Thus, all the string particles vibrate with the same frequency.

b)

The whole string lies in the same plane so that all the string particles lie in one segment. As all the string particles lie in one segment, so all of them are in the same phase.

Thus, all the particles are in the same phase.

c)

The amplitude of the wave varies from particle to particle. It is the highest at the antinode, which is twice the amplitude of the wave. The amplitude of the wave decreases gradually until it becomes zero at the nodes.

Thus, the amplitude of the points on the string is different for different points.

ii)

The equation to find the amplitude of the wave is,

A( x )=0.06sin( 2π 3 x )( 1 )

Substituting the value in the above equation, we get:

A( x )=0.06sin( 2π 3 ( 0.375 ) )( 1 ) =0.06( sin π 4 ) =0.042 m

Thus, the amplitude of the wave is 0.042 m.