# Reflection And Transmission Of Waves

## Reflection And Transmission Of Waves

When a pulse travelling along a string reaches the end, it is reflected. If the end is fixed as shown in the Fig.(14.13 a) the reflected pulse gets inverted. This can be explained using Newton’s Third Law. When the wave reaches the fixed point it exerts an upward pull on the end; according to Newton’s third law the fixed point exerts an equals and opposite force downward on the string. The result is an inverted pulse or a phase change of π.

On the other hand, if the end of the string is free to move vertically, the pulse overhoots to an amplitude twice the normal amount and the reflected pulse is not inverted, shown in
Fig.( 14.13 b). The details of the reflection process may be reconstructed by superposing the actual pulse and an imaginary pulse that approaches from the right as shown in the Fig.( 14.14). When a wave encounters a boundary between a light string and a heavy string it undergoes partial reflection and transmission.

Since the tensions are the same, the relative magnitudes of the wave velocities are determined by the mass densities.

v = √T/μ ⇒v ∝ 1/ √μ

Consider two strings of mass densities μ1 and μ2 joined together to form a single string. The respective wave velocities in the two strings will be v1 and v2 which are given by

v1/v2 = √μ2/√μ1

If ai is the amplitude of the incident wave in the medium 1, then the amplitudes of the
reflected(ar) and transmitted (at)waves in the medium 1 and 2 respectively, are given by

ar = ai (v2 - v1/v2 + v1)   (14.29)

at = ai (2v2/v2 + v1) (14.30) IMPORTANT

1. When a wave is reflected from a denser medium (i.e. μ2 < μ1 and v1 > v2) the reflected wave gets inverted or undergoes a phase change of π; while the transmitted wave in not inverted.

In the extreme situation if μ2 >> μ1, i.e. the string 2 is so heavy that it acts like a fixed boundary, then

ar = ai (v2 - v1/v2 + v1) = ai = [ v1/v- 1)/ (v1/v2 + 1 )]

since v1 >> v2 or v2/v1 ≈ 0

∴ ar = -ai

2. When a wave is reflected from a rarer medium (i.e. μ2 < μ1 or v2> v1) the reflected wave does not undergo any phase change, that is, it is not inverted.

In the extreme case if μ2 << μ1 , i.e. the string 2 is so light that it behaves like a free end, then

ar = ai (v2 - v1/v2 + v1) = ar[(1 - v1/v2 )/ (1 + v1/v2 )]

Since v2 >> v1 or v1/v2 ≈ 0

ar = +ai