Wave Changes

Properties Reflection Refraction Diffraction Resonance Resonance in String Instruments Resonance in Wind Instruments Interference: Beats Properties

REFLECTION -the turning back of a wave or at least a portion of it when it encounters an obstacle or where the waves bounce off When waves are reflected, the process of reflection has certain properties. If a wave hits an obstacle at a right angle to the surface then the wave is reflected directly backwards.

If the wave strikes the obstacle at some other angle then it is not reflected directly backwards. The angle that the waves arrives at is the same as the angle that the reflected waves leaves at. The angle that waves arrives at or is incident at equals the angle the waves leaves at or is reflected at. Angle of incidence equals angle of reflection.

Echo is produced when the sound waves are reflected from a wall or an certain object to your ears. Echoes are the reflection of sound waves

REFRACTION – the change in direction of a wave as it crosses a boundary or another medium -accompanied by a change in speed and wavelength of the waves If sound waves travel slower in the second medium, the waves will be refracted toward the normal. NORMAL -an imaginary line perpendicular to the boundary between the mediums If sound travels faster in the second medium, the waves will be refracted away from the normal.

Sound waves can also be refracted if the speed of sound changes according to their position in a medium. The waves bend toward the region of slower speed. Sounds carry farther at night than during a sunny day.

Diffraction Spreading out of waves as they pass by the edge of an obstacle or through an opening is called diffraction. - Occurs whenever a sound wave encounters an obstacle or opening - the most evident when the wave length of the sound waves is obstacle or opening. Diffraction enables you to hear a sound from around a corner, even though no straight path exists from the source of the sound to your ears.

Resonance -the reinforcing of sound -occurs when a small repeated force produces larger and larger vibrations in an object -whenever an object is able to vibrate is acted upon by a series of periodic waves having a frequency that is equal to its natural frequency of vibration -The natural frequency of vibration

The repeated force must be applied with the same frequency as the resonance frequency of the object. Resonance Frequency – the frequency which an object would vibrate naturally if disturbed in some way. The original waves and the reflected waves combine and form wave patterns that appear to stand still. Standing Waves - also known as a stationary wave - a wave that remains in a constant position.

Resonance in String Instruments Resonance increases the loudness of the sounds produced by many musical instruments. Standing waves are set up in the column of air inside the instrument. Fundamental Frequency – the lowest natural frequency of the string’s sound The wavelength of the fundamental frequency is equal to twice the length of the string.

EXAMPLE Calculate the fundamental frequency of vibration of the A-string on a guitar. Its mass is 0.30 g, 33 cm long, and is stretched with a tension of 77 N. Given: Mass of string or m = 0.30 g = 3.0 x 10 -4 kg Length of string or L = 33 cm = 0.33 m Tension (Force) or F = 77 N = 77 kg-m/s2

Resonance in Wind Instruments Wind instruments use vibrations in air columns to produce notes. The end that air is blown into is always considered an open end that must have an antinode in a standing wave. Some instruments, like flute or saxophone, have both ends of their air column open, so the only natural frequencies picked are those that allow an antinode at each end.

Interference: Beats When two sound waves of slightly different frequency are added together, the resultant sound wave will have periodically changing loudness, known as the beat. The frequency of the beat is the difference in the two frequencies, and the frequency of the resultant of the sound is the average of the two frequencies.

Beats – periodic variations of loudness - produced because the sound waves of the two tones overlap and interfere with each other. Constructive – interference of the combined waves Destructive – interference between waves Beat Frequency – number of beats per second

QUESTIONS A guitar string with a length of 80.0 cm is plucked. The speed of a wave in the string is 400 m/sec. Calculate the frequency of the first, second, and third harmonics. Answer A pitch of Middle D (first harmonic = 294 Hz) is sounded out by a vibrating guitar string. The length of the string is 70.0 cm. Calculate the speed of the standing wave in the guitar string. Answer

QUESTIONS A frequency of the first harmonic is 587 Hz (pitch of D5) is sounded out by a vibrating guitar string. The speed of the wave is 600 m/sec. Find the length of the string. Answer

Several positions along the medium are labeled with a letter Several positions along the medium are labeled with a letter. Categorize each labeled position along the medium as being a position where either constructive or destructive interference occurs.

Twin water bugs Jimminy and Johnny are both creating a series of circular waves by jiggling their legs in the water. The waves undergo interference and create the pattern represented in the diagram at the right. The thick lines in the diagram represent wave crests and the thin lines represent wave troughs. Several of positions in the water are labeled with a letter. Categorize each labeled position as being a position where either constructive or destructive interference occurs. ANSWERS

Answers: f1 = 250 Hz; f2 = 500 Hz; f3 = 750 Hz Given: L = 0.80 m v = 400 m/s Let = wavelength = 2• L = 2 • (0.80 m) = 1.6 m Now rearrange the wave equation v = f • to solve for frequency. f1 = v / = (400 m/s) / (1.6 m) = 250 Hz The frequencies of the various harmonics are whole-number multiples of the frequency of the first harmonic. Each harmonic frequency (fn) is given by the equation fn = n • f1 where n is the harmonic number and f1 is the frequency of the first harmonic. BACK

Answer: v = 412 m/s Given: L = 0.70 m f = 294 Hz (1st) The strategy for solving for the speed of sound will involve using the wave equation v = f • where is the wavelength of the wave. The frequency is stated but the wavelength must be calculated from the given value of the length of the string. For the first harmonic, the wavelength is twice the length of the string. = 2 • L = 2 • (0.70 m) = 1.4 m Now substitute into the wave equation to solve for the speed of the wave. v = f • = (294 Hz) • (1.4 m) v = 412 m/s BACK

L = 0.5 • 1.02 m = 0.51 m BACK Answer: L = 0.51 m Given: f = 587 Hz v = 600 m/s The length of a guitar string is related mathematically to the wavelength of the wave which resonates within it. Thus the strategy for solving for length will be to first determine the wavelength of the wave using the wave equation and the knowledge of the frequency and the speed. The wave equation states that v = f • where is the wavelength of the wave. Rearranging this equation and substituting allows one to determine the wavelength. L = v / f = (600 m/s) / (587 Hz) = 1.02 m For the first harmonic, the length of the string is one-half the wavelength of the wave. Thus, the following calculation can be performed: L = 0.5 • 1.02 m = 0.51 m BACK

Answers: Constructive Interference: G, J, M and N Destructive Interference: H, I, K, L, and O Constructive Interference: A and B Destructive Interference: C, D, E, and F