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The displacement of the medium in a sound wave is given by the equation y 1 = A cos (ax + bt) where A, a and b are positive constants. The intensity of the reflected wave is 0.64 times that of the incident wave. (a) -What are the wave length and frequency of incident wave? (b) Write the equation for the reflected wave. (c) In the resultant wave formed after reflection, find the maximum and minimum values of the particle speeds in the medium. (d) Express the resultant wave as a superposition of a standing wave and a travelling wave. What are the. Positions of the antinodes of the standing wave? What is the direction of propagation of travelling wave?

The displacement of the medium in a sound wave is given by the equation y1 = A cos (ax + bt) where A, a and b are positive constants. The intensity of the reflected wave is 0.64 times that of the incident wave.
(a) -What are the wave length and frequency of incident wave?
(b) Write the equation for the reflected wave.
(c) In the resultant wave formed after reflection, find the maximum and minimum values of the particle speeds in the medium.
(d) Express the resultant wave as a superposition of a standing wave and a travelling wave. What are the. Positions of the antinodes of the standing wave? What is the direction of propagation of travelling wave?

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
7 years ago
Hello Student,
Please find the answer to your question
(a) KEY CONCEPT : Use the equation of a plane progressive wave which is as follows.
y = A cos (2π/ℷ x + 2 π v t)
The given equation is y1 = A cos (ax + bt)
On comparing, we get 2 π/ℷ = a ⇒ ℷ = 2 π/a
Also, 2 πv = b
⇒ v = b/2 π
(b) Since the wave is reflected by an obstacle, it will suffer a phase difference of π. The intensity of the reflected wave is 0.64 times of the incident wave.
Intensity of original wave I ∝ A2
Intensity of reflected wave I’ = 0.64 I
⇒ I’ ∝ A’2 ⇒ 0.64 I ∝ A’2
⇒ 0.64 A2 ∝ A’2 ⇒ A’ ∝ 0.8A
So the equation of resultant wave becomes
y2 = 0.8A cos (ax – bt + π) = -0.8A cos (ax – bt)
(c) KEY CONCEPT : The resultant wave equation can be found by superposition principle
y = y1 + y2 = A cos (ax + bt) + [-0.8 A cos ( ax – bt) ]
The particle velocity can be found by differentiating the above equation
v = dy/dt = - Ab sin(ax + bt) – 0.8 Ab sin (ax – bt)
= -Ab [sin (ax + bt) + 0.8 sin (ax – bt)]
= -Ab [sin axcos bt + cos ax sin bt + 0.8 sin ax cos bt – 0.8 cos ax sin bt]
v = -Ab [ 1.8 sin ax cos bt + 0.2 cos ax sin bt]
The maximum velocity will occur when sin ax = 1 and cos bt = 1 under these condition cos ax = 0
And sin bt = 0
∴ |vmax| = 1.8 Ab
Also, |vmin| = 0
(d) y = [A cos (ax + bt)] – [0.8 A cos (ax – bt)]
= [0.8 A cos (ax + bt) + 0.2 A cos (ax + bt)] – [0.8A cos (ax – bt)]
= 0.8 A [-2 sin {(ax + bt) + (ax – bt)/2} Sin {(ax + bt) – (ax – bt)/2}] 0.2 A cos (ax + bt)]
⇒ y = -1.6 A sin ax sin bt + 0.2 A cos (ax + bt)
Where (-1.6 A sin ax sin bt ) is the equation of travelling wave.
The wave is travelling in –x direction.
NOTE : Antinodes of the standing waves are the positions where the amplitude is maximum,
i.e. sin ax = 1 = sin [nπ + (-1)n π/2 ]
⇒ x = [n + (-1)n/2] π/a

Thanks
Aditi Chauhan
askIITians Faculty

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