Wave practice Longitudinal all sound waves are longitudinal

Wave practice

Longitudinal: all sound waves are longitudinal Longitudinal: waves that have compressions and rarefactions are longitud

Transversal: perpendicular (up and down) displacement. Longitudinal: waves that have compressions and rarefactions are longitudinal

Transversal: perpendicular (up and down) displacement.

Transversal: perpendicular (up and down displacement.

1. Draw a wave diagram and show the following: amplitude of 5 meters and a wavelength of 10 meters. 5 m 10 m

2. Determine the period of a wave that has a frequency to 14 hertz. T = 1/f T = 1/14 seconds T = 0. 0714 seconds

3. Determine the frequency that has a period of 4 seconds. f = 1/T f = ¼ Hz f = 0. 25 Hz

4. 16 m 10 m 4 m 5 m 15 m 25 m Determine the wavelength and the amplitude of the wave. Amplitude: 6 m wavelength: 20 m

5. Knowing that the velocity of a wave is determined by the equation v = f l, a. calculate the velocity if the wave has a frequency of 12 Hz and a wavelength of 3 meters. b. calculate the frequency if the velocity is 8 m/s and the wavelength is 4 meters. c. calculate the wave length if the frequency is 16 hertz and the velocity is 8 m/s. d. calculate the velocity of the wave if the period of the wave is 2 seconds and the wave length is 4 meters.

a. calculate the velocity if the wave has a frequency of 12 Hz and a wavelength of 3 meters. v = f l v = 12 x 3 = 36 m/s

b. calculate the frequency if the velocity is 8 m/s and the wavelength is 4 meters. v = f l 8 = f (4) f = 2 Hz

c. calculate the wave length if the frequency is 16 hertz and the velocity is 8 m/s. v = f l 8 = 16 l l = 0. 5 m

d. calculate the velocity of the wave if the period of the wave is 2 seconds and the wave length is 4 meters. v = f l v = 1/T l v = ½ (4) v = 2 m/s

6. A B Explain which of the two waves has a higher frequency. Wave B has a higher frequency because it has a shorter wavelength.

1 Jerome and Claire are doing the Period of a Pendulum Lab. They observe that a pendulum makes exactly 10 complete back and forth cycles of motion in 21. 8 seconds. Determine the period of the pendulum. Period = how long it takes to complete one cycle Period = time/cycles = 21. 8 seconds / 10 cycles = 2. 18 seconds /cycle = 2. 18 seconds

2 Strong winds can apply a significant enough force to tall skyscrapers to set them into a back-and-forth motion. The amplitudes of these motions are greater at the higher floors and barely observable for the lower floors. It is said that one can even observe the vibrational motion of the Sears Tower in Chicago on a windy day. As the Sears Tower vibrates back and forth, it makes about 8. 6 vibrations in 60 seconds. Determine the frequency and the period of vibration of the Sears Tower. Frequency = vibrations / seconds = 8. 6 / 60 = 0. 1433 vibrations/seconds = 0. 1433 Hz Period = seconds / vibrations = 60 / 8. 6 = 6. 98 seconds/vibrations

3. Extreme waves along ocean waters are sometimes referred to as rogue waves. Merchant ships have reported rogue waves which are estimated to be 25 m high and 26 m long. Assuming that these waves travel at 6. 5 m/s, determine the frequency and period of these waves. v = f l f = v / l f = 6. 5 / 26 = 0. 25 Hz T = 1/f = 1/0. 25 = 4 seconds

4 4. Tsunamis are much different than rogue waves. While rogue waves and other waves are generated by winds, tsunamis originate from geological events such as movements of tectonic plates. Tsunamis tend to travel very fast. A tsunami generated off the coast of Chile in 1990 is estimated to have traveled approximately 6200 miles to Hawaii in 15 hours. Determine the speed in mi/hr and m/s. (Given: 1. 0 m/s = 2. 24 mi/hr) v = d/t = 6200 miles / 15 hours = 413. 33 miles / hr 413. 33 / 2. 24 = 184. 52 m/s

5 Microbats use echolocation to navigate and hunt. They emit pulses of high frequency sound waves which reflect off obstacles and objects in their surroundings. By detecting the time delay between the emitted pulse and the return of the reflected pulse, a bat can determine the location of the object. Determine the time delay between the sending of a pulse and the return of its reflection from an object located 12. 5 m away. Approximate the speed of the sound waves as 345 m/s. Total distance traveled = 2 x 12. 5 = 25 m d = vt t = d/v = 25 m / 345 m/s = 0. 0725 seconds

6. Susie is listening to her favorite radio station - 102. 3 FM. The station broadcasts radio signals with a frequency of 1. 023 x 108 Hz. The radio wave signal travel through the air at a speed of 3. 0 x 108 m/s. Determine the wavelength of these radio waves. v = f l l = v / f = 3. 0 x 108 / 1. 023 x 108 = 2. 93 m

7. A transverse wave is observed to be moving along a lengthy rope. Adjacent crests are positioned 2. 4 m apart. Exactly six crests are observed to move past a given point along the medium in 9. 1 seconds. Determine the wavelength, frequency and speed of these waves. Wavelength = 2. 4 meters Total distance = (6)(2. 4) = 14. 4 m Speed: S = d/t = 14. 4 m / 9. 1 s = 1. 58 m/s Frequency = 6 waves /9. 1 seconds = 0. 659 Hz Or V = l f 1. 58 = 2. 4 f f = 1. 58/2. 4 = 0. 659 Hz

8. A wave is traveling in a rope. The diagram below represents a snapshot of the rope at a particular instant in time. Determine the number of wavelengths which is equal to the horizontal distance between points: a. C and E 1 b. C and K 3. 5 c. A and J 4 d. B and F 1. 5 e. D and H 2 f. E and I 1. 75

9. A wave with a frequency of 12. 3 Hz is traveling from left to right across a rope as shown in the diagram at the right Positions A and B in the diagram are separated by a horizontal distance of 42. 8 cm. Positions C and D in the diagram are separated by a vertical distance of 12. 4 cm. Determine the amplitude, wavelength, period and speed of this wave. Amplitude = CD / 2 = 12. 4 /2 = 6. 2 cm = 0. 062 m Period = 1/frequency = 1 / 12. 3 = 0. 0813 seconds Since there are two waves from A to B, the wave length is the distance from A to B divided by 2. wavelength = 42. 8 /2 = 21. 4 cm or 0. 214 m. Speed = wavelength x frequency = 0. 214 x 12. 3 = 2. 63 m/s

10. A rope is held tightly and shook until the standing wave pattern shown in the diagram at the right is established within the rope. The distance A in the diagram is 3. 27 meters. The speed at which waves move along the rope is 2. 62 m/s. a. Determine the frequency of the waves creating the standing wave pattern. Wavelength = 3. 27 / 1. 5 (each antinode = 0. 5 wavelength) = 2. 18 m V = l f f = V / l = 2. 62 / 2. 18 = 1. 2 Hz

A rope is held tightly and shook until the standing wave pattern shown in the diagram at the right is established within the rope. The distance A in the diagram is 3. 27 meters. The speed at which waves move along the rope is 2. 62 m/s. b. Determine the number of vibrational cycles which would be measured in 20. 0 seconds. frequency = 1. 2 vibration / seconds in 20 seconds, we have 1. 2 x 20 = 24 vibrations

11. In a physics demonstration, Mr. H establishes a standing wave pattern in a snakey by vibrating it up and down with 32 vibrations in 10 seconds. Gerald is holding the opposite end of the snakey and is standing 6. 2 m from Mr. H's end. There are four equal length sections in the snakey, each occupied by an antinode. Determine the frequency, wavelength and speed of the wave. Frequency = vibrations / second = 32 / 10 = 3. 2 Hz 4 equal length = 4 antinodes = 4 x (0. 5 wavelengths) = 2 wavelengths There are 2 wavelengths in 6. 2 meters 1 wavelength in 3. 1 meters V = l f = 3. 1 x 3. 2 = 9. 92 m/s

12. A standing wave pattern is established in a 246 -cm long rope. A snapshot of the rope at a given moment in time is shown in the diagram below. Vibrations travel within the rope at speeds of 22. 7 m/s. Determine the frequency of vibration of the rope. 3 wavelengths in 246 cm 1 wavelength in 246/3 cm 82 cm = 0. 82 m V = l f f = V / l f = 22. 7 / 0. 82 = 27. 68 Hz