ch15-p050 - 50. Consider that the length of the spring as...

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2 2 2(0.00296 J) 1.81 rad/s 0.0018 kg m m K I ω == = . Therefore the angular frequency of the oscillation is = m / θ o = 34.6 rad/s. Using Eq. 15-5, then, the period is T = 0.18 s. 50. Consider that the length of the spring as shown in the figure (with one of the block’s corners lying directly above the block’s center) is some value L (its rest length). If the (constant) distance between the block’s center and the point on the wall where the spring attaches is a distance r , then r cos = d / 2 and r cos = L defines the angle measured from a line on the block drawn from the center to the top corner to the line of r (a straight line from the center of the block to the point of attachment of the spring on the wall). In terms of this angle, then, the problem asks us to consider the dynamics that results from increasing from its original value o to o + 3º and then releasing the system and letting it oscillate. If the new (stretched) length of spring is L (when = o + 3º), then it is a straightforward trigonometric exercise to show that ( L ) 2 = r 2
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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