OldMidterm4 - oldmidterm 04 PAPAGEORGE MATT Due 4:00 am...

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oldmidterm 04 – PAPAGEORGE, MATT – Due: Apr 29 2008, 4:00 am 1 Question 1, chap 15, sect 1. part 1 of 1 10 points Hint: Write down equations for x ( t ) and v ( t ) and use sin 2 + cos 2 = 1 to calculate ω . A mass attached to a spring executes sim- ple harmonic motion in a horizontal plane with an amplitude of 0 . 337 m. At a point 0 . 27297 m away from the equilibrium, the mass has speed 3 . 32 m / s. What is the period of oscillation of the mass? Correct answer: 0 . 374014 s (tolerance ± 1 %). Explanation: There are several ways to solve this prob- lem — for example, one may use the energy conservation — but the simplest solution uses nothing but the equation of simple harmonic motion x ( t ) = A sin( ω t ) , and its time derivative v ( t ) = d x dt = A ω cos( ω t ) . For any angle [such as the phase angle ω t ] sin 2 ( ω t ) + cos 2 ( ω t ) = 1 , hence parenleftBig x A parenrightBig 2 + parenleftBig v A ω parenrightBig 2 = 1 , and therefore parenleftBig v ω parenrightBig 2 = A 2 x 2 . Consequently, ω = v A 2 x 2 = 16 . 7993 s 1 . Finally, the period of the oscillation follows from the angular frequently, T = 2 π ω = 0 . 374014 s . Question 2, chap 15, sect 3. part 1 of 1 10 points A block of unknown mass is attached to a spring of spring constant 9 . 92 N / m and undergoes simple harmonic motion with an amplitude of 9 . 02 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be v = 39 . 2 cm / s. Calculate the mass of the block. Correct answer: 0 . 393925 kg (tolerance ± 1 %). Explanation: Basic Concepts Energy conservation: If K is kinetic energy and U is potential energy, K i + U i = K f + U f Kinetic energy of particle with mass m and speed v : K = 1 2 m v 2 Mass m on spring with constant k : ω = radicalbigg k m and potential energy of a spring at displace- ment x : U = 1 2 k x 2 Period T = 2 π ω Solution: Call the maximum displacement (amplitude) A . The halfway displacement is A/ 2. Energy conservation requires 0 + 1 2 k A 2 = 1 2 m v 2 + 1 2 k parenleftbigg A 2 parenrightbigg 2 or k A 2 = m v 2 + 1 4 k A 2 so m = 3 k A 2 4 v 2 = (3) (9 . 92 N / m) (0 . 0902 m) 2 (4) (0 . 392 m / s) 2 = 0 . 393925 kg .
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oldmidterm 04 – PAPAGEORGE, MATT – Due: Apr 29 2008, 4:00 am 2 Question 3, chap 11, sect 3. part 1 of 1 10 points A thin hoop of radius 2 . 56 m and mass 9 . 5 kg is suspended from a pivot on the hoop itself as shown in the figure. θ The acceleration of gravity is 9 . 8 m / s 2 . Find the period T of small oscillations of this hoop around θ = 0. Correct answer: 4 . 54153 s (tolerance ± 1 %). Explanation: The hoop oscillates as a physical pendulum. Its center of mass is in its geometric center, at distance R = 2 . 56 m from the pivot. So when it swings by a small angle θ away from the equilibrium position (center directly be- low the pivot), there is a torque due to gravity force, τ = m g R sin θ m g r θ (1) for small θ . The hoop has moment of inertia I 0 = m R 2 around its center of mass. The momend of inertia relative to the pivot follows via the parallel axis theorem I = I 0 + m R 2 = 2 m R 2 . (2) Consequently, the equation of motion for the swinging hoop I d 2 θ dt 2 = τ (3) becomes 2 m R 2 d 2 θ dt 2 = m g R θ , (4) which simplifies to d 2 θ dt 2 = g 2 R θ . (5)
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