# hw31 - Overton Mays Homework 31 Due 4:00 am Inst Turner...

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Overton, Mays – Homework 31 – Due: Apr 20 2005, 4:00 am – Inst: Turner 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points When an object oscillating in simple harmonic motion is at its maximum displacement from the equilibrium position, which of the follow- ing is true of the values of its speed and the magnitude of the restoring force? Magnitude of Speed Restoring Force 1. Zero Maximum correct 2. Maximum Zero 3. Zero Zero 4. 1 2 maximum 1 2 maximum 5. Maximum 1 2 maximum Explanation: The maximum displacement occurs at the turning points, which are the points where the velocity or speed is zero. The magnitude of the restoring force is given by Hooke’s law F = - k x, where k is the spring constant and x is the displacement. Since x is a maximum, F is a maximum. From a different perspective, the displace- ment from the equilibrium position can be written as y = A sin θ , where θ is the phase of the oscillation. When the object is at its maximum dis- placement, θ = π 2 . The speed is then v = ω A cos θ = ω A cos π 2 = 0 . The restoring force is F = m A ω 2 sin θ = m A ω 2 π 2 = m A ω 2 , at its maximum value. 002 (part 1 of 1) 10 points A block rests on a flat plate that executes vertical simple harmonic motion with a period of 1 . 09 s. The acceleration of gravity is 9 . 8 m / s 2 . What is the maximum amplitude of the motion for which the block does not separate from the plate? Correct answer: 0 . 29493 m. Explanation: Basic Concepts: Newton’s 2 nd law: X F = m a Simple harmonic motion: x = A cos( ω t ) a = d 2 x dt 2 = - A ω 2 cos( ω t ) . Solution: At each instant, there are two forces acting on the block: the force of gravity F g = m g and the normal force N from the plate. The block separates from the plate when N = 0. Newton 2 nd law reads - m g + N = m a , where up is positive. We see that N = m ( g + a ) so for N to vanish, we must have a max = - g , (1) (the normal force N also vanishes for any acceleration quicker than this, but we are only looking for the point when the block just barely leaves the plate). We also know that for simple harmonic motion a = - A ω 2 cos( ω t ) ,

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Overton, Mays – Homework 31 – Due: Apr 20 2005, 4:00 am – Inst: Turner 2 and at the highest and lowest points, the ac- celeration is at a maximum. The acceleration of the block is positive (up) at the lowest point, and negative (down) at the highest point. We want the negative one, so a max = - A max ω 2 . Comparing with equation (1), we note A max ω 2 = g , or A max = g ω 2 , and using T = 2 π ω , we find A max = g T 2 4 π 2 = (9 . 8 m / s 2 ) (1 . 09 s) 2 4 π 2 = 0 . 29493 m . 003 (part 1 of 1) 10 points A block of unknown mass is attached to a spring of spring constant 9 . 8 N / m and undergoes simple harmonic motion with an amplitude of 15 . 2 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be v = 34 . 8 cm / s.
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