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Amplitude of reflected and transmitted wave on a string

Amplitude of reflected and transmitted wave on a string

Do you know about the law of conservation of energy? According to it, Energy can not be created and destroyed. It remains conserved for an isolated system. Applying it to the reflection of waves through the partially fixed and partially Free End. When a wave reaches the boundary, the energy associated with it also conserves and will be equal to the energy of reflected and transmitted waves.Lets see it in detail !

Table of content

  • Introduction
  • Derivation
  • Practice problem
  • FAQs

Introduction

As we know, when a wave on string reaches the junction of a rare and denser medium, some of its part is transmitted and some reflected. As wave carriers energy and energy is proportional to the square of amplitude. If the energy is divided between reflected and transmitted waves, the amplitude of reflected and transmitted waves will be lesser than the amplitude of incident wave. Let's derive the formula for the amplitude of reflected and transmitted waves.

Derivation :

Consider that the incident wave is on the light string (rarer medium) and its amplitude is Ai. Assume that the speed of the incident wave and transmitted wave is v1 and v2, respectively. If the incident wave is transmitted to the heavy string (denser medium), then v1>v2.

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At the boundary, A part of the incident wave gets reflected and a part of it gets transmitted. Let the amplitude of the reflected wave and transmitted wave be Ar and At .

Considering the propagation direction of the incident wave along the positive x-direction, And equation is given as

yi=Ai sin ωt-k1x.....(i)

Since the incident wave is along the positive x-direction, the reflected and transmitted waves propagate along the negative x-direction and the positive x-direction, respectively. Hence,

The equation of the reflected wave is given by,

yr=Ar sin ωt+k1x.....(ii)

As the reflected wave travels in the same medium, the velocity and the wave number will be the same as the incident wave.

The equation of the transmitted wave is given by,

yt=At sin ωt-k2x.....(iii)

Now for the continuity of displacement at the boundary (x=0)

yi+yr=yt

Putting the values from equation (i) , (ii) and (iii)

 

Ai sin ωt+Ar sin ωt=At sin ωtAi+Ar=At...(iv)

 

Also, at the boundary the slope of the wave will be continuous, i.e.

yix+yrx=ytx

From equation (i) , (ii) and (iii)

 

yix=-Ai k1cos ωt-k1x yrx=Ar k1cos ωt-k1x

 

 

-Ai k1cos ωt+Ar k1cos ωt=-At k2cos ωtAi-Ar=k2k1 At

 

Ai-Ar=k2k1 At

As k=v

Ai-Ar=v1v2 At...(v)

Adding question (iv) and (v)

 

2Ai=1+v1v2AtAt=2 v2v2+v1 Ai  ......(vi)

 

Now subtracting question (v) from equation (iv), we get,

 

2Ar=1-v1v2At2Ar=v2-v1v22 v2v2+v1 AiAr=v2-v1v2+v1 Ai....(vii)

 

Hence eq (vii) gives the amplitude of reflected wave and eq (vi) gives the amplitude of the transmitted wave. You can observe from the equations that both the amplitude is less than the amplitude of the incident wave.

Practice problem

Q.The equation of a plane progressive wave is given by yi=0.6 sin 2 t-x2 . On reflection from a denser medium, its amplitude becomes 23 of the amplitude of the incident wave. Which of the following represents the equation of the reflected wave?

  1. yr=0.6 sin 2π t+x2
  2. yr=-0.4 sin 2π t+x2
  3. yr=0.4 sin 2π t+x2yr=-0.4 sin 2π t-x2
     

A. Since the wave is traveling from a rarer medium to a denser medium, it gets inverted, i.e., the amplitude has a negative value. Then, we can represent the equation of a reflected wave as follows:

 

yr=-Ar sin 2π t+x2As,Ar=23 AiAr=23×0.6=0.4

 

So the equation is given as

yr=-0.4 sin 2π t+x2

Thus, option (B) is the correct answer.

Q. The amplitude of the incident wave on a string is 10 cm. What is the amplitude of the transmitted wave, given the velocity of wave in light string is 450 cm/s and in heavy string is 150 cm/s.

A. The amplitude of the transmitted wave is

 

At=2 v2v1+v2 AiAt=2 ×150500+150×10At=2 ×150450+150×10At=0.5×10At=5 cmAns

 

Q. If the ratio of velocities of wave in rarer medium to the denser medium is 83 What is the ratio of amplitude of reflected to incident waves?

A. Given v1v2=83

As we know the amplitude of reflected wave is given as

 

Ar=v2-v1v2+v1 AiOr,ArAi=1-v1v21+v1v2ArAi=1-831+83ArAi=-53113ArAi=-53113ArAi=-511Ans

 

Q. A wave traveling on a light string gets reflected from a heavy string. If the velocity in a light string is 240 m/s and in a heavy string is 80 m/s, find the fraction of energy transmitted.

A. the amplitude of the transmitted wave is

 

At=2 v2v1+v2 AiAt=2 ×80240+80 AiAt=0.5 Ai

 

As energy is proportional to the square of amplitude

 

EA2Et=(0.5)2 EiEt=0.25 Ei

 

Hence 0.25 part of the incident energy will get transmitted.

FAQs

Q. Amplitude of reflected and transmitted waves on a string depends upon ?
A.
It depends on the velocities of the wave in the rarer and denser medium.

Q. What is the reflection coefficient?
A.
The ratio of amplitude of reflected wave to the amplitude of incident wave is known as the reflection coefficient.

Q. What is the transmission coefficient?
A.
The ratio of amplitude of the transmitted wave to the amplitude of the incident wave is known as the transmission coefficient.

Q. What is the sum of the reflection coefficient and transmission coefficient?
A.
As both are the fraction of the incident wave, hence the sum is equal to 1.

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