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0DD32z5wbaYN_1210016299_jwk572

# 0DD32z5wbaYN_1210016299_jwk572 - oldmidterm 04 – KIM JI...

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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 Δ x into the water is Δ F = − ( ρ l w Δ x ) g = ( m l + m r ) d 2 (Δ x ) dt 2 = − ( m l + m r ) ( ω 2 Δ x ) ω 2 = ρ l w g m r + m l ω = radicalBigg ρ l w g m r + m l 2 π f = radicalBigg ρ l w g m r + m l f = 1 2 π radicalBigg ρ l w g m r + m l = 1 2 π 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 μ = m/L , and the tension on the string is T = μv 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 it’s 1 . 465 m long. The second pipe is open at one end and closed at the other end; it’s 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 = λ 2 , 2 λ 2 , 3 λ 2 , 4 λ 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 = λ 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 = λ 4 + λ 2 × an integer = λ 4 , 3 λ 4 , 5 λ 4 , 7 λ 4 , . . ....
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0DD32z5wbaYN_1210016299_jwk572 - oldmidterm 04 – KIM JI...

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