The transverse displacement of a string (clamped at its both ends) is given by
`y(x, t) = 0.06 sin ((2 x)/(3) x) cos (120 pi t)`
where x and y are in m and t in s. The length of the string is 1.5 m and its mass is `3.0 xx 10^(-2) kg`.
Answer the following :
(a) Does the function represent a travelling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency , and speed of each wave ?
(c ) Determine the tension in the string.

Options

A. same frequency
B. same phase
C. same energy
D. different amplitude

Answer

A, B, D

Explanation

For given stationary wave, following is an appropriate figure. (When N = node or nodal point, A = anti node or antinodal point).

(a) All the particles oscillate with same fre quency. i.e. no. of oscillations completed by all the particles in unit time is same. Hence, option (A) is correct. (b) Between two consecutive nodes, all par ticles either simultaneously displace upwards or downwards. Thus, during oscillations their phases remain same. Hence, option (B) is true.
(c) Energy of a given particle of a string (in which stationary waves are produced) is directly proportional to its amplitude. But amplitude of a given particle in stationary wave depends on its position along the length of the string, according to formula A. = 2a sin(kx). Hence, all the particles between two consecutive nodes will not have equal amplitude and so they will not have same energy. Hence, option (C) is false.
(d) Between two consecutive nodes, amplitudes of all the particles are not equal. Hence, option (D) is true.