WAVES AND SOUND AP PHYSICS 1

TYPES OF WAVES Transverse waves- Particles move perpendicular to the waves motion Longitudinal Waves (compression waves)- Particles move parallel to the waves motion Compressions (molecules are close together) Rarefactions (molecules are far apart)

WAVELENGTH AND SPEED

EXAMPLE #1 Sound waves travel in air with a speed of 343 m/s. The lowest frequency sound we can hear is 20.0 Hz; the highest frequency is 20.0 kHz. Find the wavelength of sound for these frequencies.

EXAMPLE #2 A 5.0m length of rope, with a mass of 0.52kg, is pulled taut with a tension of 46N. Find the speed of waves on the rope.

EXAMPLE #3 A 12-m rope is pulled tight with a tension of 92N. When one end of the rope is given a “thunk” it takes 0.45s for the disturbance to propagate to the other end. What is the mass of the rope?

EXAMPLE #4 A rope of length L and mass M hangs from a ceiling. If the bottom of the rope is given a gentle wiggle, a wave will travel to the top of the rope. As the wave travels upward does its speed: a)Increase b)Decrease c)Stay the same

WAVE REFLECTIONS There are two possibilities of reflections a wave encounters. Fixed Reflections are inverted. Free or Open End Reflections are not inverted.

EXAMPLE #5 Five seconds after a brilliant flash of lightning, thunder shakes the house. Was the lightning… a)About a mile away b)Much closer than a mile away c)Much farther away than a mile EXAMPLE #6 You drop a stone from rest into a well that is 7.35m deep. How long does it take before you hear the splash?

SOUND INTENSITY EXAMPLE #7 A loudspeaker puts out 0.15W of sound through a square area of 2.0m on each side. What is the intensity of this sound?

POINT SOURCE EXAMPLE #8 Two people relaxing on a deck listen to a songbird sing. One person, only 1.00m from the bird, hears the sound with an intensity of 2.80 x W/m 2. a)What intensity is heard by the second person, who is 4.25m from the bird? Assume that no reflected sound is heard by either person. b)What is the power output of the bird’s song?

INTENSITY VS LOUDNESS

EXAMPLE # 9 A crying child emits sound with an intensity of 8.0 x W/m 2. Find: a)The intensity level in decibels for the child’s sounds b)The intensity level for this child and its twin both crying with identical intensities.

DOPPLER EFFECT The change in frequency due to relative motion between a source and a receiver.

SUPERPOSITION EXAMPLE # 10

INTERFERENCE Constructive Interference: Wave that add to give a larger amplitude. Destructive Interference: Waves that add to give a smaller amplitude. Interference Patterns: Waves that overlap can create patterns of constructive and destructive interference. In Phase/Opposite (Out) Phase: Two sources are in phase if they both emit crests at the same time. They are opposite phase if one emits a crest at the same time as another emits a trough.

EXAMPLE #11 Two speakers separated by a distance of 4.30m emit sound of frequency 221 Hz. The speakers are in phase with one another. A person listens from a location 2.80m directly in front of one of the speakers. Does this person hear constructive or destructive interference? EXAMPLE #12 Two speakers separated by a distance of 5.20m emit sound of frequency 104 Hz. The speakers are out of phase with one another. A person stands 3m in front of one of the speakers and 1.30m to one side of the center line between them. What type of interference occurs at the person’s location?

HARMONICS A vibrating string will produce standing waves whose frequencies depend upon the length of the string.

In the lowest frequency of vibration, one wavelength will equal twice the length of string and its called the fundamental frequency (f 1 ). For f 1, 1λ = 2L One wavelength = 2*length of string Fundamental FrequencyHalf of a wavelength

HARMONICS A Harmonic series is a series of frequencies that include the fundamental frequency and multiples of that frequency. 1st harmonic = f 1 2 nd harmonic = f 2 = 2*f 1 3 rd harmonic = f 3 = 3*f 1 Etc…

HARMONICS The second harmonic is the next possible standing wave for the same string length. This shows an increase in frequency, and a decrease in wavelength. f 2 =2f 1 λ 2 = L Second Harmonic = 2*fundamental frequency

HARMONICS As the harmonic increases the frequency increases and wavelength decreases. Ex: f 3 = 3f 1 λ 3 = 2/3λ 1 f 4 = 4f 1 λ 4 = ½ λ 1

FORMULA FOR OTHER HARMONICS

EXAMPLE # 13 One of the harmonics on a string 1.30m long has a frequency of Hz. The next higher harmonic has a frequency of Hz. Find: a)The fundamental frequency b)The speed of waves on this string

STANDING WAVES IN AN AIR COLUMN If one end of the pipe is closed, only odd harmonics are present (1, 3, 5, etc). This changes the formula: f n = n* V n=1, 3, 5… 4L Frequency = harmonic number*(speed of waves on in the pipe) (4)*length of vibrating air column)

EXAMPLE # 14 An empty soda bottle is to be used as a musical instrument in a band. In order to be tuned properly the fundamental frequency of the bottle must be Hz. a)If the bottle is 26cm tall, how high should it be filled with water to produce the desired frequency? b)What is the frequency of the next higher harmonic for this bottle?

STANDING WAVES IN AN AIR COLUMN If both ends of a pipe are open, all harmonics are present and the ends act as antinodes. This is the exact opposite of a vibrating string, but the waves act the same so we can still use the same formula to calculate frequencies. fn = n* V n=1, 2, 3… 2L Frequency = harmonic number x (speed of waves on the in the pipe) (2)*length of vibrating air column)

EXAMPLE # 15 If you fill your lungs with helium and speak, you sound something like Donald Duck. From this observation we can conclude that the speed of sound in helium must be: a)Less than the speed of sound in air b)The same as the speed of sound in air c)Greater than the speed of sound in air

EXAMPLE What are the first three harmonics in a 2.45 m long pipe that is open at both ends? Given that the speed of sound in air is 345 m/s. L= 2.45 m v= 345 m/s f n = n*v/2L 1 st harmonic: f 1 = 1*(345 m/s)/(2*2.45 m) = 70.4 Hz 2 nd harmonic: f 2 = 2*(345 m/s)/(2*2.45 m) = 141 Hz 3 rd harmonic: f 3 = 3*(345 m/s)/(2*2.45 m) = 211 Hz

EXAMPLE What are the first three harmonics of this pipe when one end of the pipe is closed? Given that the speed of sound in air is 345 m/s. L= 2.45 m v= 345 m/s f n = n*v/4L 1 st harmonic: f 1 = 1*(345 m/s)/(4*2.45 m) = 35.2 Hz 3 rd harmonic: f 3 = 3*(345 m/s)/(4*2.45 m) = 106 Hz 5 th harmonic: f 5 = 5*(345 m/s)/(4*2.45 m) = 176 Hz

BEATS