0DD32z5wbaYN_1210016299_jwk572

0DD32z5wbaYN_1210016299_jwk572 - oldmidterm 04 KIM, JI Due:...

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Unformatted text preview: oldmidterm 04 KIM, JI Due: Jan 14 2008, 10:00 am 1 Question 1, chap 16, sect 2. part 1 of 1 10 points A 4 m by 10 m raft whose mass is 16000 kg floats in fresh water. A 6060 kg load is dropped on the raft. The acceleration of gravity is 9 . 8 m / s 2 . With what frequency does the raft bob up and down? Correct answer: 0 . 670904 Hz (tolerance 1 %). Explanation: Given : l = 10 m , w = 4 m , m r = 16000 kg , and m l = 6060 kg . The force exerted when the raft drops x into the water is F = ( l w x ) g = ( m l + m r ) d 2 ( x ) dt 2 = ( m l + m r ) ( 2 x ) 2 = l w g m r + m l = radicalBigg l w g m r + m l 2 f = radicalBigg l w g m r + m l f = 1 2 radicalBigg l w g m r + m l = 1 2 braceleftbigg (1000 kg / m 3 ) (10 m) (4 m) (9 . 8 m / s 2 ) 16000 kg + 6060 kg bracerightbigg 1 / 2 = 0 . 670904 Hz Question 2, chap 16, sect 2. part 1 of 1 10 points Transverse waves with a speed of 64 . 8 m / s are to be produced in a taut string. A 4 . 64 m length of string with a total mass of 0 . 0363 kg is used. What is the required tension? Correct answer: 32 . 8503 N (tolerance 1 %). Explanation: The mass per unit length of the string is = m/L , and the tension on the string is T = v 2 = mv 2 L = 32 . 8503 N . Question 3, chap 16, sect 3. part 1 of 1 10 points Consider two organ pipes. The first pipe is open at both ends and its 1 . 465 m long. The second pipe is open at one end and closed at the other end; its 2 . 226 m long. When both pipes are played together, the first overtone the lowest harmonic above the fundamental frequency of the second pipe produces beats agains the fundamental harmonic of the first pipe. What is the fre- quency of these beats? Take the sound speed in air to be 331 m / s. Correct answer: 1 . 44637 Hz (tolerance 1 %). Explanation: The first pipe is open at both ends, so a resonating standing wave in it has pressure nodes at both ends. This implies L o = 2 , 2 2 , 3 2 , 4 2 , . . . and therefore f = v 2 L o , 2 v 2 L o , 3 v 2 L o , 4 v 2 L o , . . . For the fundamental harmonic of the pipe L o = 2 = f fund o = v 2 L o . Numerically, f fund o = 331 m / s 2(1 . 465 m) = 112 . 969 Hz . oldmidterm 04 KIM, JI Due: Jan 14 2008, 10:00 am 2 The second pipe is open at one and and closed at the other. A resonating standing wave in this pipe has a pressure node at the open end and a displacement node at the closed end. A displacement node is an antin- ode of pressure and lies quarter-wavelength from the nearest pressure node. Hence, the distance between the two ends of the string must be L co = 4 + 2 an integer = 4 , 3 4 , 5 4 , 7 4 , . . ....
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This note was uploaded on 05/05/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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0DD32z5wbaYN_1210016299_jwk572 - oldmidterm 04 KIM, JI Due:...

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