Reflection and Refraction of Sound Wave

Reflection of sound wave

Take two metal tubes A and B. Keep one end of each tube on a metal plate. Place a wrist watch at the open end of the tube A and interpose a cardboard between A and B. Now at a particular inclination of the tube B with the cardboard, ticking of the watch is clearly heard. The angle of reflection made by the tube B with the cardboard is equal to the angle of incidence made by the tube A with the cardboard.

Applications of reflection of sound waves

(i) Whispering gallery :

The famous whispering gallery at St. Paul’s Cathedral is a circular shaped chamber whose walls repeatedly reflect sound waves round the gallery, so that a person talking quietly at one end can be heard distinctly at the other end. This is due to multiple reflections of sound waves from the curved walls.

(ii) Stethoscope :

Stethoscope is an instrument used by physicians to listen to the sounds produced by various parts of the body. It consists of a long tube made of rubber or metal. When sound pulses pass through one end of the tube, the pulses get concentrated to the other end due to several reflections on the inner surface of the tube. Using this doctors hear the patients’ heart beat as concentrated rays.

(iii) Echo :

Echoes are sound waves reflected from a reflecting surface at a distance from the listener. Due to persistence of hearing, we keep hearing the sound for 101th of a second, even after the sounding source has stopped vibrating. Assuming the velocity of sound as 340 ms–1, if the sound reaches the obstacle and returns after 0.1 second, the total distance covered is 34 m. No echo is heard if the reflecting obstacle is less than 17 m away from the source.

Refraction of sound

This is explained with a rubber bag filled with carbon-di-oxide as shown in Fig. 7.9. The velocity of sound in carbon-di-oxide is less than that in air and hence the bag acts as a lens. If a whistle is used as a source S, the sound passes through the lens and converges at O which is located with the help of flame. The flame will be disturbed only at the point O.

When sound travels from one medium to another, it undergoes refraction.

Refer this video to know more about reflection and refraction of sound

Applications of refraction of sound

It is easier to hear the sound during night than during day-time. During day time, the upper layers of air are cooler than the layers of air near the surface of the Earth. During night, the layers of air near the Earth are cooler than the upper layers of air. As sound travels faster in hot air, during day-time, the sound waves will be refracted upwards and travel a short distance on the surface of the Earth. On the other hand, during night the sound waves are refracted downwards to the Earth and will travel a long distance.

The reflection of the sound follows the law angle of incidence equals to the angle of reflection, called the law of reflection. The incident , the reflected and the normal wave all lie in the same plane. When a longitudinal sound wave strikes a flat surface, sound is reflected in a coherent manner provided that the dimension of the reflective surface is large compared to the wavelength of the sound. Sound waves propagate into a medium by the vibration of an object. Sound waves travel in straight line until something interferes with their path.  The change in speed makes the sound wave to bend away from its original direction. The change in angle or direction of a sound wave is called Refraction. When waves travel from one medium to another then due to the variation in the speed there is a change in the direction of propagation and this phenomenon is called Refraction. The Speed of sound is fast in warm air and is slow in cold air. Sun tends to warm the upper layer of air sometimes and cold at the down layer of air. The sound waves may bend a little due to this, called sound refraction.

Problem (JEE Main):

Speed of transverse wave in a string of density 100 kg/m³ and area of cross-section 10 mm² under a tension of 10³ N is,

(a) 100 m/s         (b) 1000 m/s

(c) 200 m/s        (d) 2000 m/s

Solution:

We know that, v = √T/m

Here, m = mass per unit length = ρs

So, v = √T/ρs

= √(1000/1001010^-6)

= 10³ m/s

Therefore, from the above observation we conclude that, option (b) is correct.