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Unformatted text preview: Chapter 13 solved problems. Some of the numbers are slightly different from the book but the solution (procedure) is the same. 14. Picture the Problem : The mass is attached to the spring and pulled down from equilibrium and released. The stiffness of the spring causes the mass to oscillate at the given frequency. Strategy: Since the mass starts at x = A at time t = 0, this is a cosine function given by ( 29 cos x A t = . From the data given you need to identify the constants A and . Solution: 1. Identify the amplitude as A : A = 6.4 cm 2. Calculate from the period: 2 2 0.83 s T = = 3 . Substitute A and into the cosine equation: ( 29 2 6.4 cm cos 0.83 s x t = Insight: A cosine function has a maximum amplitude at t = 0, but a sine function has zero amplitude at t = 0. That is why we chose a cosine function to describe the motion. 18. Picture the Problem : A mass is attached to a spring. The mass is displaced from equilibrium and released from rest. The spring force causes the mass to oscillate about the equilibrium position in harmonic motion. Strategy: Since the mass starts from rest at t = 0, the harmonic equation will be a sine function. We will use the amplitude and period to determine the general equation, which can be evaluated for any specific time. During the first half of each period the mass will be moving in the negative x-direction toward the minimum and during the second half the mass will move in the positive direction, back toward the maximum. Therefore we can determine the direction of motion by finding in which half of a period the time is located....
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This note was uploaded on 01/05/2012 for the course CHEM 3321 taught by Professor Bean during the Spring '11 term at American Jewish University.
- Spring '11