The transverse displacement of a string (clamped at its both ends) is given by
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 ×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.
Given:
…(1)
(a) The general equation of an standing wave is given by,
y(x, t) = 2 sin kx × cos wt …(2)
The equation 1, is similar to equation 2, hence the given equation represents a standing wave.
(b) A wave travelling along the positive x-direction is given by,
y1 = a sin(wt-kx)
For a wave travelling in negative x direction is given by,
y2 = a sin (wt + kx)
The superposition of these two waves can be written as,
y = y1 + y2
y = a sin (wt-kx) – a sin(wt + kx)
The terms can be expanded using the formula of sin(a + b) and the result is,
y = 2 a sin(kx) cos(wt)
y = 2a sin (2πx/λ) cos(2πvt) …(3)
By comparing equation (1) and (3) we have,
2 π/λ = 2 π/3
⇒ wavelength, λ = 3 m
It is given that,
120 π = 2 πv
Frequency, v = 60Hz
Wave speed, V is given by,
V = v× λ
V = 60 s-1× 3m
V = 180 ms-1
(c) The velocity of a transverse wave in a string is given by,
V = (T/m)1/2
Where,
V = velocity of wave,
T = tension in string
m = mass per unit length of string
⇒ m = mass of the string / length of the string
⇒ m = 3.02× 10-2Kg/ 1.5 m
⇒ m = 2× 10-2 Kgm-1
Tension in the string, T = V2m
T = 2× 10-2Kgm-1× (180ms-1)2
T = 648 N