Version HW# - Vibrations & Waves arts (4) This print-out should have 5 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Superposition. points You are given two waves, a transverse wave thatmovestotherightf (x)andatransverse wave that moves to the left f (x), on a string. Astheproblembegins,thewavef (x)ismoving to the right at v = + m/s and the wave f (x) is moving to the left at v = m/s. - - v v 4 5 6 7 8 9 What is the shape of the wave on the string after s?.. - - - - 4 5 6 7 8 9 4 5 6 7 8 9. - - 4 5 6 7 8 9 correct 4. 5. 6. - - - - - - 4 5 6 7 8 9 4 5 6 7 8 9 4 5 6 7 8 9
7. 8. 9.. - - - - - - - - Version HW# - Vibrations & Waves arts (4) 4 5 6 7 8 9 4 5 6 7 8 9 4 5 6 7 8 9 4 5 6 7 8 9 - - Initial time, t = s 4 5 6 7 8 9 After s the positions of the two waves are have both moved meters in opposite directions. The sum of the two wave is the resultant wave, the light gray line. Superpostion, at t = s - - 4 5 6 7 8 9 - - Resultant, at t = s 4 5 6 7 8 9 The initial wave moving to the right is represented withadashed lineandtheeavemoving to the left is represented with a dotted line. Vibrating Wire. points Whenaparticularwireisvibratingwithafrequency of. Hz, a transverse wave of wavelength 46.6 cm is produced. Determine the speed of wave pulses along the wire.
Version HW# - Vibrations & Waves arts (4) Correct answer:.578 m/s. Let : f =. Hz and λ = 46.6 cm. If the frequency is f and the wavelength is λ then the speed of wave is v = f λ = (. Hz)(46.6 cm) =.578 m/s. m cm The tensions are the same, so µ v = µ v µ v = µ v v = v = (8 m/s) =.7 m/s. Serway CP 49 4 (part of ). points Tension is maintained in a string as in the figure. The observed wave speed is m/s when the suspended mass is.8 kg. Serway CP 48. points A string is 4 cm long and has a mass of g. A wave travels at 8 m/s along this string. A secondstringhasthesamelengthbuthalfthe mass of the first. If the two strings are under the same tension, what is the speed of a wave along the second string? Correct answer:.7 m/s. Let : l = 4 cm, v = 8 m/s, and m = g. Because l = l and m = m, the densities are related by µ = m l = m l = µ. The velocity of the waves is v = F µ F = µv..8 kg What is the mass per unit length of the string? The acceleration of gravity is 9.8 m/s. Correct answer:.56694 kg/m. Let : v = m/s, m =.8 kg, and g = 9.8 m/s. The tension in the string is and its velocity is F v = µ, and µ = F v = mg v F = mg = (.8 kg)( 9.8 m/s ) ( m/s) =.56694 kg/m.
Version HW# - Vibrations & Waves arts (4) 4 5 (part of ). points What is the wave speed when the suspended mass is.6 kg? Correct answer:.997 m/s. The velocity is F v = = Let : m =.6 kg. mg µ µ = (.6 kg)(9.8 m/s ).56694 kg/m =.997 m/s. Longitudinal Spring Pulses 6. points As your hand moves back and forth to generate longitudinal pulses in a spiral spring, your hand completes.6 back-and-forth cycles every 6. s. The velocity of the pulse in the spring is.98 cm/s. What is the wavelength? Correct answer:.8455 m. Frequence is the number of cycles generated per unit time, so f = n t Longitudinal Spring Pulses 7. points As your hand moves back and forth to generate longitudinal pulses in a spiral spring, your hand completes.8 back-and-forth cycles every 4.78 s. The velocity of the pulse in the spring is.978 cm/s. What is the wavelength? Correct answer:.89 m. Frequence is the number of cycles generated per unit time, so f = n t The velocity of the wave is defined by λ = vt n v = f λ = n t λ = (.978 cm/s)(4.78 s).8 =.89 m. m cm Conceptual 4 Q 8 (part of 4). points Consider two waves traveling through the same medium in the same time frame. A B The velocity of the wave is defined by v = f λ = n t λ λ = vt n = (.98 cm/s)(6. s).6 =.8455 m. m cm Compare the wavelengths.. Cannot be determined. A and B have the same wavelength.. A has the longer wavelength. correct
Version HW# - Vibrations & Waves arts (4) 5 4. B has the longer wavelength. A exhibits three complete wavelengths in the same time that B exhibits five complete wavelengths, so A has a longer wavelength. 9 (part of 4). points Compare the amplitudes.. Cannot be determined. A has the smaller amplitude.. B has the smaller amplitude. correct 4. A and B have the same amplitude. The vertical distance between the troughs and peaks of A is greater, so it has the larger amplitude. (part of 4). points Compare the frequencies.. B has the higher frequency. correct. A and B have the same frequency.. A has the higher frequency. 4. Cannot be determined v = λf f = v λ. Since the speeds are the same and A has a longer wavelength λ, then A must have the lower frequency. (part 4 of 4). points Compare the periods.. A and B have the same period.. A has the shorter period.. B has the shorter period. correct 4. Cannot be determined. T =. Since A has a lower frequency, its f period must be larger. Conceptual 4 Q (part of 4). points Consider two waves traveling through the same medium in the same time frame. A B Compare the wavelengths.. Cannot be determined. A and B have the same wavelength.. A has the longer wavelength. correct 4. B has the longer wavelength. A exhibits three complete wavelengths in the same time that B exhibits five complete wavelengths, so A has a longer wavelength. (part of 4). points Compare the amplitudes.. B has the smaller amplitude. correct. A and B have the same amplitude.. A has the smaller amplitude. 4. Cannot be determined The vertical distance between the troughs and peaks of A is greater, so it has the larger amplitude.
Version HW# - Vibrations & Waves arts (4) 6 4 (part of 4). points Compare the frequencies.. A has the higher frequency.. B has the higher frequency. correct. Cannot be determined 4. A and B have the same frequency. v = λf f = v λ. Since the speeds are the same and A has a longer wavelength λ, then A must have the lower frequency. 5 (part 4 of 4). points Compare the periods.. A and B have the same period.. B has the shorter period. correct. Cannot be determined. 4. A has the shorter period. T =. Since A has a lower frequency, its f period must be larger.