Incident, Reflected and Transmitted waves at a boundary
If a pulse is introduced at the left end of the rope, it will travel through the rope towards the right end of the medium. This pulse is called the incident pulse since it is incident towards (i.e., approaching) the boundary with the pole. When the incident pulse reaches the boundary, two things occur: 1. A portion of the energy carried by the pulse is reflected and returns towards the left end of the rope. The disturbance that returns to the left after bouncing off the pole is known as the reflected pulse.
2. A portion of the energy carried by the pulse is transmitted to the pole, causing the pole to vibrate.
Write the equation of reflected wave for a given incident wave at a rigid boundary
Example: Find the phase change between incident and reflected sound wave from a fixed wall by writing equations of incident and reflected wave.
Solution: When a wave is reflected from a wall, the wave is inverted . As the phase difference between a wave and its inverted wave is nˉ, the phase difference is π y=Acos(kx+ωt) yr=−Acos(kx+ωt) =Acos(kx+ωt+π) ⇒ϕ=π
Reflection of sound
Sound waves (like any other wave) return back in the same medium obeying the laws of reflection. The only requirement for the reflection of the sound wave is that the reflecting surface must be bigger than the wavelength of the wave. Reflection of sound wave is seen in echoes, megaphone, ear trumpet etc.
Explain refraction of sound
Refraction of sound waves is most evident in situations in which the sound wave passes through a medium with gradually varying properties. For example, sound waves are known to refract when traveling over water. Even though the sound wave is not exactly changing media, it is traveling through a medium with varying properties; thus, the wave will encounter refraction and change its direction.
Equation of reflected wave for a given incident sound wave
Example: The displacement of the medium in a sound wave is given by the equation; y1=Acos(ax+bt) where a & b are positive constants. The wave is reflected by an obstacle situated at x = 0 . The intensity of the reflected wave is 0.64 times that if the incident wave. The equation for the reflected wave is given as y2=±10sAcos(ax−bt) Find s
Solution: equation for sound intensity is I=21ρω2A2v Ir=0.64Ii 21ρω2Ar2v=20.64ρω2Ai2v Ar2=0.64Ai2 Ar=0.8Ai=0.8A 10sA=0.8A s=8
Reflection of a given incident sound wave in an open tube
Example: A train of sound waves is propagated along a wide pipe and it is reflected from an open end. If the amplitude of the waves is 0.002 cm, the frequency 1000 Hz and the wavelength 40 cm, Find the amplitude of vibration at a point 10 cm from open end inside the pipe.
Wavelength is 40cm Amplitude to be found at 10 cm =440cm=4λ In open pipe from open pipe at a distance of λ/4 always a node is formed. So, amplitude at 10 cm from open end will be zero because amplitude at node is zero.
Incident, reflected and transmitted sound waves at a boundary
The simplest situation of reflection and transmission occurs when waves are impinging normal to the surface. The case of a longitudinal wave incident on the interface between two media is shown in the above figure.
Laws of reflection of sound
Law 1 : Incident wave,normal to the reflecting surface and reflected wave at the point of incidence all lie in the same plane. Law 2: Angle of incidence is equal to the angle of reflection.