Review Problems Test 3 F

Version 001 – HW 8 Bonus – spurlock – (41003) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. HoltSF12C01 001 10.0points A mass of 0.27 kg is attached to a spring and is set into vibration with a period of 0.39 s. What is the spring constant of the spring? Correct answer: 70 . 08 N / m. Explanation: Let : m = 0 . 27 kg and T = 0 . 39 s . T = 2 π radicalbigg m k parenleftbigg T 2 π parenrightbigg 2 = m k k = m parenleftbigg 2 π T parenrightbigg 2 = (0 . 27 kg) parenleftbigg 2 π 0 . 39 s parenrightbigg 2 = 70 . 08 N / m . OscillationonaSpring 002(part1of3)10.0points A 494 g mass is connected to a light spring of force constant 3 N / m that is free to oscillate on a horizontal, frictionless track. The mass is displaced 9 cm from the equilibrium point and released from rest. 3 N / m 494 g 9 cm x = 0 x Find the period of the motion. Correct answer: 2 . 54966 s. Explanation: Let : m = 494 g = 0 . 494 kg and k = 3 N / m . This situation corresponds to the special case x ( t ) = A cos ωt , so ω = radicalbigg k m = radicalBigg 3 N / m 0 . 494 kg = 2 . 46432 s − 1 and the period is T = 2 π ω = 2 π 2 . 46432 s − 1 = 2 . 54966 s . 003(part2of3)10.0points What is the maximum speed of the mass? Correct answer: 0 . 221789 m / s. Explanation: Let : A = 9 cm . The velocity as a function of time is v ( t ) = − ω A sin( ω t ) , so the maximum speed of the mass is v max = ω A = (2 . 46432 s − 1 ) (0 . 09 m) = 0 . 221789 m / s . 004(part3of3)10.0points What is the maximum acceleration of the mass? Correct answer: 0 . 546559 m / s 2 . Explanation: The acceleration as a function of time is a ( t ) = − ω 2 A cos( ω t ) ,

Version 001 – HW 8 Bonus – spurlock – (41003) 2 so the maximum acceleration of the mass is a max = ω 2 A = (2 . 46432 s − 1 ) 2 (0 . 09 m) = 0 . 546559 m / s 2 . TiplerPSE51444 005 10.0points A 22 kg person steps into a car of mass 2513 kg, causing it to sink 6 . 65 cm on its springs. Assuming no damping, with what fre- quency will the car and passenger vibrate on the springs? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 179988 Hz. Explanation: Let : m = 22 kg , M = 2513 kg , g = 9 . 8 m / s 2 , and Δ x = 6 . 65 cm = 0 . 0665 m . The spring force is F = k Δ x k = F Δ x = m g Δ x , where m is the person’s mass. The frequency is f = 1 2 π radicalbigg k m + M = 1 2 π radicalbigg m g ( m + M ) Δ x = 1 2 π radicalBigg (22 kg) (9 . 8 m / s 2 ) (22 kg + 2513 kg) (0 . 0665 m) = 0 . 179988 Hz . HoltSF12B01 006 10.0points You need to know the height of a tower, but darkness obscures the ceiling. You note that a pendulum extending from the ceiling almost touches the floor and that its period is 23 s. The acceleration of gravity is 9 . 81 m / s 2 . How tall is the tower? Correct answer: 131 . 451 m. Explanation: BasicConcept: T = 2 π radicalBigg L g Given: T = 23 s g = 9 . 81 m / s 2 Solution: parenleftbigg T 2 π parenrightbigg 2 = L g L = g parenleftbigg T 2 π parenrightbigg 2 = (9 . 81 m / s 2 ) parenleftbigg 23 s 2 π parenrightbigg 2 = 131 . 451 m TiplerPSE51430 007 10.0points Military specifications often call for electronic devices to be able to withstand accelerations of 10 g . To make sure that their products meet this specification, manufacturers test them using a shaking table that can vibrate a device at various specified frequencies and amplitudes. If a device is given a vibration of amplitude 5 . 9 cm, what should be its frequency in order to test for compliance with the 10 g military specification? The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 6 . 48976 Hz. Explanation:

Version 001 – HW 8 Bonus – spurlock – (41003) 4 Explanation: Let : f max = 4200 Hz and v = 334 m / s . λ min = v f max = 334 m / s 4200 Hz = 0 . 0795238 m . HoltSF12Rev55 012 10.0points Yellow light travels through a certain glass block at a speed of 1 . 97 × 10 8 m/s. The wavelength of the light in this particular type of glass is 3 . 81 × 10 − 7 m (381 nm). What is the frequency of the yellow light in the glass block? Correct answer: 5 . 1706 × 10 14 Hz. Explanation: Let : v = 1 . 97 × 10 8 m / s and λ = 3 . 81 × 10 − 7 m . f = v λ = 1 . 97 × 10 8 m / s 3 . 81 × 10 − 7 m = 5 . 1706 × 10 14 Hz . Concept20P09 013 10.0points A grunting porpoise emits a sound of 61 Hz sound. What is the wavelength of this sound in water, where the speed of sound is 1500 m / s? Correct answer: 24 . 5902 m. Explanation: Let : v = 1500 m / s and f = 61 Hz . v = f λ λ = v f = 1500 m / s 61 Hz = 24 . 5902 m . Conceptual1505 014 10.0points Anna was on vacation and came across an echo lake. Wanting to know how far she had to swim to get across the lake to the other side, she yelled across “Hello!” How wide is the lake if 7 seconds later she heard her own echo? Assume that the tem- perature is 20 ◦ C and the speed of sound is 344 m / s. Correct answer: 1204 m. Explanation: Let : v = 344 m / s and t = 3 . 5 . It took 3 . 5 seconds for the sound to reach the other side, so the width of the lake is d = v t = (344 m / s) (3 . 5 s) = 1204 m . AmbulanceDrivervsCarDriverp2 015 10.0points An ambulance is traveling east at 47 m / s. Behind it there is a car traveling along the same direction at 31 . 5 m / s. The ambulance driver hears his siren at a frequency of 798 Hz. 31 . 5 m / s Car 47 m / s Ambulance At what frequency does the driver of the car hear the ambulance’s siren? The velocity of sound in air is 343 m / s . Correct answer: 766 . 285 Hz. Explanation:

Version 001 – HW 8 Bonus – spurlock – (41003) 5 Let : v = 343 m / s , v car = 31 . 5 m / s , and f 0 = 798 Hz . From the Doppler shift, f ′ = v sound ± v observer v sound ∓ v source f . The car moves toward the ambulance and the ambulance moves away from the car, so f ′ = v + v car v + v amb f 0 = parenleftbigg 343 m / s + 31 . 5 m / s 343 m / s + 47 m / s parenrightbigg (798 Hz) = 766 . 285 Hz . FallingTuningFork 016 10.0points A tuning fork vibrating at 256 Hz falls from rest and accelerates at 9 . 8 m / s 2 . How far below the point of release is the tuning fork when waves of frequency 216 . 2 Hz reach the release point? The speed of sound in air is 340 m / s . Correct answer: 238 . 362 m. Explanation: Let : v s = 340 m / s , g = 9 . 8 m / s 2 , f 1 = 256 Hz , and f 2 = 216 . 2 Hz . From the Doppler effect, the change of the frequency is f 2 = f 1 v s v s + v f 2 ( v s + v ) = f 1 v s v = v s parenleftbigg f 1 f 2 − 1 parenrightbigg , so the time of the fall is t fall = v g = v s g parenleftbigg f 1 f 2 − 1 parenrightbigg = 340 m / s 9 . 8 m / s 2 parenleftbigg 256 Hz 216 . 2 Hz − 1 parenrightbigg = 6 . 38675 s . The distance the fork falls during this time is d fall = g t 2 fall 2 ; the fork continues to fall until the sound re- turns: t return = d fall v s = g t 2 fall 2 v s = (9 . 8 m / s 2 ) (6 . 38675 s) 2 2 (340 m / s) = 0 . 587865 s , so the total time the fork falls is t total = t fall + t return = 6 . 38675 s + 0 . 587865 s = 6 . 97462 s and the total distance fallen is d total = g t 2 total 2 = (9 . 8 m / s 2 ) (6 . 97462 s) 2 2 = 238 . 362 m . FrequencyFromaPlane 017 10.0points An airplane traveling at one third the speed of sound ( i.e. , 172 m / s) emits a sound of frequency 7 . 06 kHz. At what frequency does a stationary lis- tener hear the sound as the plane approaches? Correct answer: 10 . 59 kHz. Explanation: Let : v = 172 m / s , and f = 7 . 06 kHz . The plane approaches the listener, so f ′ = v v − v 3 f = 1 . 5 f = 1 . 5 (7 . 06 kHz) = 10 . 59 kHz .