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# Waves 2

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### Waves 2

1. 1. Waves -2 See more at:  Facebook – https://www.facebook.com/AdityaAbeysinghePr esentations  Slideshare - slideshare.net/adityaabeysinghe  Wordpress adityaabeysinghepresentations.wordpress.com/ abeysinghe-foundation/ Waves-2 By Aditya Abeysinghe 1
2. 2. Wave Properties Waves have various properties. Properties of waves depend on the type of wave and the mode of energy transmission. However, all these characteristics of different waves can be studied under 4 main topics: 1. Reflection 2. Refraction 3. Diffraction 4. Interference Waves-2 By Aditya Abeysinghe 2
3. 3. Reflection Like light waves, mechanical waves (transverse and longitudinal) and sound can also be reflected at a surface. As in light waves, the reflection of any other type of wave also obeys the laws of reflection: 1. Angle of incidence is equal to the angle of reflection E.g.: Source Receiver Waves-2 By Aditya Abeysinghe 3
4. 4. Sound waves come to a focus when they are incident on a curved concave mirror. Also, the sound waves are distributed over large distances when the sound source is kept on the focus of a curved surface. E.g.: Loudspeaker Waves-2 By Aditya Abeysinghe 4
5. 5. Refraction Before understanding how waves can be refracted let’s find out how refraction occurs. Refraction of any source of energy is due to the change in speed of the waves on entering a different medium. E.g.: Sound may propagate on air faster than water due to various factors, but diffrence in density been the major factor. All energy sources that refract off a surface obey the laws of refraction. Waves-2 By Aditya Abeysinghe 5
6. 6. Consider the following diagram. Incident ray i – incident angle r – refraction angle i Normal r Refracted ray Laws of refraction: 1. The incident and refracted rays, and the normal at the point of incidence, all lie in the same plane. 2. For two given media, sin i /sinr is a constant. (i is the angle of incidence and r is the angle of refraction. Waves-2 By Aditya Abeysinghe 6
7. 7. Examples: 1. The refraction of sound explains why sounds are easier to hear at night than during day-time. Day time: Night time: Upper layers of the air are colder than the layers near the ground level Upper layers of the air are warmer than the layers near the ground level Ground level Ground level Sound waves travel faster the higher the temperature. So, during the day time the sound waves are refracted away from the ground level. Thus, the sound intensity decreases. But, during night, sound waves are refracted towards the ground level and the intensity of sound increases. Waves-2 By Aditya Abeysinghe 7
8. 8. 2. Effect of wind on sound Consider two occasions when, (i) The wind is blowing in the direction of sound. wind Sound Source Observer Here the bottom of the sound wavefront is moving more slowly than the upper part. So, the wavefront turns towards the observer (O). So, the sound is heard easily. (ii) The wind is blowing away from the direction of the sound. Sound wind Source Observer Here the upper part of the sound wavefront is moving more slowly than the bottom part. So, the wavefront turns away from the observer (O). So, the sound intensity diminishes. Waves-2 By Aditya Abeysinghe 8
9. 9. Diffraction Diffraction occurs when any type of wave meets an obstacle. The behavior of waves on facing an obstacle varies depending on the size of the obstacle. When the size of the aperture in the obstacle incresase, there is less pressure on that aperture, so the wave maintains a low diffraction pattern beyond the obstacle and viceversa. Diffraction can be clearly observed using a ripple tank. Waves-2 By Aditya Abeysinghe 9
10. 10. Ripple tank - Waves-2 By Aditya Abeysinghe 10
11. 11. The diffraction bands or the patterns of the diffracted wave fronts can be illustrated as follows: Narrow slit More diffraction Wide slit Less diffraction Less diffraction Diffraction Long waves Short waves Waves-2 By Aditya Abeysinghe 11
12. 12. The reason for such a variation in the wave patterns is due to the change in wavelength. Generally, the smaller the width of the aperture in relation to the wavelength, the greater is the diffraction. A similar effect can be observed when sound waves are diffracted at a surface. Diffraction increases with longer wavelength. By the inverse relationship between the frequency (f) and the wave length (λ), it is clear as to why sounds with high frequency defract less. Waves-2 By Aditya Abeysinghe 12
13. 13. Interference When two or more waves of the same frequency overlap, interference occurs. Interference can be easily explained using the theory of superposition. Consider the diagram below. If the two A B Waves-2 By Aditya Abeysinghe oscillators are in phase, crests from A will meet crests from B and troughs from B will meet troughs from B. Similarly when the two oscillators are not in phase, troughs/crests from A will meet crests/troughs from B. 13
14. 14. waves. Applying the Principle of superposition Consider two occasions: (i) When the two waves are in phase (ii) When the two waves are not in phase Waves-2 By Aditya Abeysinghe 14
15. 15. When the two waves are in phase Crests from wave A will meet crests from Wave B and troughs from wave A will meet troughs from wave B. This process increases the amplitude of the resultant wave. Thus, this process is called the constructive interference A Trough Trough When the two waves are not in phase Crest Crest B Waves-2 By Aditya Abeysinghe Trough Crests from wave A will meet troughs from Wave B and troughs from wave A will meet crests from wave B. This process decreases the amplitude of the resultant wave. Thus, this process is called the destructive interference 15
16. 16. Now consider what happens when two oscillators are placed and activated on a ripple tank. A bright spot can be observed due to the increase of the amplitude of the resultant wave. Thus, white bands are due to constructive interference. Similarly, dark spot can be observed due to the decrease of the amplitude of the resultant wave. Thus, dark bands are due to the destructive interference. Waves-2 By Aditya Abeysinghe 16
17. 17. Interference of sound wavesConsider the following diagram. X Destructive interference Constructive interference Destructive interference Constructive interference Destructive interference As the ear or microphone is moved along the line XY, a larger or a smaller sound is heard depending on the type of interference (constructive or destructive) Constructive interference Destructive interference Y Waves-2 By Aditya Abeysinghe 17
18. 18. Example: Consider two loudspeakers 2m apart. Both of these loudspeakers are connected to the same oscillator and both emit a sound of 850Hz. A sensitive detector moving parallel to the XY line, along JK, detects a maximum wave at J and another at K. (JK = 1m) Assume that the distance between Y and K is 4.8m. Calculate the speed of sound waves in the air. X J 2m 1m K Y 4.8 m Waves-2 By Aditya Abeysinghe 18
19. 19. Since both at J and at K, sound can be heard, J and K can be considered to be points of constructive interference. Furthermore, XJ = YJ. Since, K is the next point at which a sound was heard, XK – YK = λ (λ is the wavelength) XK can be found using the Pythagoras theorem, XK = √ XY2 + YK2 = 5.2 m Therefore, λ = XK – YK = 5.2m – 4.8m = 0.4m Thus, the velocity of sound on air, V = fλ V = 850 × 0.4 = 340 ms-1 19 Waves-2 By Aditya Abeysinghe