105 - final exam - Fall 2009 - solutions

105 - final exam - Fall 2009 - solutions - Instructions:...

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Instructions: Record your answers on the bubble sheet. The Testing Center no longer allows students to see which problems they got right & wrong, so I strongly encourage you to mark your answers in this test booklet . You will get this test booklet back (but only if you write your CID at the top of the first page ). You may write on this exam booklet, and are strongly encouraged to do so. In all problems, ignore friction, air resistance, and the mass of all springs, pulleys, ropes, cables, strings etc., unless specifically stated otherwise. Use g = 9.8 m/s 2 only if there are “9.8” numbers in the answer choices; otherwise use g = 10 m/s 2 . Problem 1. In the “ladies belt demo” (the belt was like a “closed-closed” string), the fundamental frequency is seen at 400 Hz. What frequency will have five antinodes? a. 800 Hz b. 1000 c. 1200 d. 1600 e. 2000 f. 2400 Hz 1. Closed-closed: Harmonic number is same as number of antinodes. f n = nf 1 (n = 1, 2, 3, …) = 5*400 = 2000 Hz. Choice E. Problem 2. Two students play with an extra-long Slinky. The student on the left end sends waves to the other student by shaking her end back and forth. After the waves die down, both students take a step backwards and try it again. How will the speed of the waves now compare to the previous waves? a. They will go faster b. They will go slower c. They will go the same speed 2. More tension faster wave speed. Choice A Problem 3. Two tuning forks have frequencies of 440 and 450 Hz. How many beats per second will you hear when they are sounded simultaneously? a. 2 beats/sec b. 4 c. 5 d. 10 e. 15 f. 20 beats/sec 3. f beat = |f 1 – f 2 | = 450 – 440 = 10 Hz. Choice D. Problem 4. A mass on a frictionless surface attached to a horizontal spring oscillates with an amplitude of 4 cm. If the spring constant is 200 N/m and the object has a mass of 0.500 kg, determine the maximum speed of the object. a. Less than 0.35 m/s b. 0.35 – 0.45 c. 0.45 – 0.55 d. 0.55 – 0.65 e. 0.65 – 0.75 f. 0.75 – 0.85 g. More than 0.85 m/s 4. conservation of energy: ½ kx 2 = ½ mv 2 v 2 = kx 2 /m v = sqrt(k/m)*x = sqrt(200/0.5) * 0.04 = sqrt(400) * 0.04 = 20*0.04 = 0.8 m/s. Choice F 1
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Problem 5. The fundamental frequency (first harmonic) of the trumpet I brought to class is very close to 150 Hz. How long would its uncoiled length be? Take the speed of sound in air to be 300 m/s. (Hint: a trumpet is like an open-open pipe.) a. Less than 0.45 m b. 0.45 – 0.55 c. 0.55 – 0.65 d. 0.65 – 0.75 e. 0.75 – 0.85 f. 0.85 – 0.95 g. More than 0.95 m 5. open-open, for fundamental freq: L = ½ λ (think of picture) = ½ (v/f) = ½ (300)/150 = ½*2 = 1 m. Choice G Problem 6. A man wants to know the height of a tower, so he cleverly sets up a long pendulum extending from the top of the tower to the ground. He measures the period of the pendulum to be exactly 10 s. How tall is the tower? a.
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105 - final exam - Fall 2009 - solutions - Instructions:...

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