Physics Laboratory Experiments
Physics Laboratory Experiments
8th Edition
ISBN: 9781285738567
Author: Jerry D. Wilson, Cecilia A. Hernández-Hall
Publisher: Cengage Learning
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Chapter 4, Problem 2EP
To determine

The physical parameters that can be varied in the simple pendulum arrangement.

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If you did the previous question right, you hopefully got an expression for yo. You may notice that you can simplify the differential equation a little bit: d'y k (y – yo) dt2 т The parameter yo now plays the roll of the "relaxed length". A better term may be "equilibrium value for y". But mathematically, it's identical to a relaxed length with the spring as the only force. We continue using this equation: y(t) = Y0 + A cos(wt + y) Now, solve for A (in cm) with these parameters. Again, if you need more information, enter -100000. The parameters are: •m = 200 grams • Yo = (equilibrium value) = 40 cm • k = (spring constant) = 0.03 N/cm
Problems 5 & 6 refer to the mass-spring oscillator depicted in the figure on the right. The block has a mass of 350 g, and is attached to a spring with a spring constant of k pushed 20 cm to left from C, its equilibrium position, before being released and allowed to move horizon- tally on a frictionless surface. . Ax = 20 cm 45 N. The mass is initially m A B DE 5. What is the frequency of oscillation for the block? A. 0.13 Hz В. 0.35 Hz С. 1.8 Hz D. 7.8 Hz 5. 6. What is the speed of the block as it passes point B? A. 0.50 m/s В. 1.0 m/s C. 1.5 m/s D. 2.0 m/s 6.
Our unforced spring mass model is mx00 + βx0 + kx = 0 with m, β, k >0. We know physically that our spring will eventually come to rest nomatter the initial conditions or the values of m, β, or k. If our modelis a good model, all solutions x(t) should approach 0 as t → ∞. Foreach of the three cases below, explain how we know that both rootsr1,2 =−β ± Sqrt(β^2 − 4km)/2mwill lead to solutions that exhibit exponentialdecay.(a) β^2 − 4km > 0. (b) β^2 − 4km =0. (c) β^2 − 4km >= 0.
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