phyexam4_pdf

# phyexam4_pdf - Version 098 – Midterm 4 – florin...

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Unformatted text preview: Version 098 – Midterm 4 – florin – (58140) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A particle hangs from a spring and oscillates with a period of 0 . 4 s. With Particle Without Particle If the mass-spring system remained at rest, by how much would the mass stretch it from its normal equilibrium position? The acceler- ation of gravity is 9 . 8 m / s 2 . 1. . 317743 m 2. . 0794358 m 3. . 124777 m 4. . 635486 m 5. . 158872 m 6. . 0397179 m correct Explanation: Let : T = 0 . 4 s and g = 9 . 8 m / s 2 . When the particle hangs and oscillates on the spring, we have T = 2 π radicalbigg m k k = 4 π 2 m T 2 . At equilibrium with the particle hung, we have k x = mg x = mg k = g T 2 4 π 2 = (9 . 8 m / s 2 )(0 . 4 s) 2 4 π 2 = . 0397179 m . 002 10.0 points It is impossible for two particles executing simple harmonic motion to remain in phase with each other if they have different 1. masses. 2. periods. correct 3. restoring forces. 4. amplitudes. 5. kinetic energies. Explanation: θ = ω t + θ = 2 π T t + θ Therefore, if the periods of the simple har- monic motion are different, the two particles can’t remain in phase. 003 10.0 points Simple harmonic motion can be described us- ing the equation x = x m sin( ωt + φ ) . If x = initial position, v = initial velocity, then 1. tan φ = − ω x v 2. tan φ = + v ω x 3. tan φ = − v ω x 4. tan φ = + ω x v correct Version 098 – Midterm 4 – florin – (58140) 2 Explanation: x = x m sin( ω t + φ ) v = dx dt = x m ω cos( ω t + φ ) When t = 0, x = x m sin φ v = x m ω cos φ x m sin φ x m ω cos φ = x v tan φ = ω x v . 004 10.0 points A large block with mass 26 kg executes horizontal simple harmonic motion as it slides across a frictionless surface with a frequency . 63 Hz . A smaller block with mass 3 kg rests on it, as shown in the figure, and the coefficient of static friction between the two is μ s = 0 . 465 . k 26 kg μ s = 0 . 465 3 kg What maximum amplitude of oscillation can the system have if the top block is not to slip? The acceleration of gravity is 9 . 8 m / s 2 . 1. 4.71522 2. 2.11539 3. 7.03015 4. 12.442 5. 1.51749 6. 2.03477 7. 29.0829 8. 8.00711 9. 18.8569 10. 4.00382 Correct answer: 29 . 0829 cm. Explanation: Let : m b = 26 kg , m t = 3 kg , f = 0 . 63 Hz , μ s = 0 . 465 , and g = 9 . 8 m / s 2 . f s mg N For simple harmonic motion, we have x = A cos( ω t + φ ) and a = − Aω 2 cos( ω t + φ ) , where ω = 2 π f = 2 π (0 . 63 Hz) = 3 . 95841 rad / s and since the extreme values of the cosine are ± 1, a max = Aω 2 . The force of friction is f s ≤ μN ....
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phyexam4_pdf - Version 098 – Midterm 4 – florin...

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