17-20 - Simple Harmonic Motion Find a function that modelsthe simple harmonic motion having the given properties. Assumethat the displacement is at its maximum at time i = 0.•17. amplitude 60 ft, period 0.5 min18. amplitude 35 cm, period 8 s19. amplitude 2.4 m, frequency 750 Hz20. amplitude 6.25 in., frequency 60 Hz

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College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Final Answer must be decimal or whole number only!!!
Adapt the method of Example 6 to determine the amplitude, period, and phase shift for the given function. Graph the function over one period indicating the x- intercepts and the coordinates of the highest and lowest points on the graph. (a) y = 3 cos(7 - 2x) amplitude period phase shift y y 3- 3 3 3 5/7 Зя 4 4 4 4 4 2 o-3 y 4. 4 2 x-intercepts (х, у) - (smaller x-value) (х, у) - (larger x-value) highest points (x, y) = (smaller x-value) (x, y) = (larger x-value) lowest point (x, y) =
3. The displacement equation of an object in simple harmonic motion is 3π π given by x(t) = 5.00 mm cos (3t+) where t is in seconds. Find the 4 speed of the object.
If the frequency of the motion of a simple harmonic oscillator is increased by a factor of 4, by what factor does the maximum speed of the oscillator change? 1/4 O It does not change. 04 02 O 1/2
1. Given the equation and graph shown below, what is the amplitude (A), period (T), angular frecuency (w), and phase qifference ($)? Harmonic motion, y(t) = 25 * sin(n/3 *t + T/4) 30 10 -10 -20 -30 0. 2n 3n 4m 5m Time, s 20 Displacement, cm
Pls help
15-22 need solution
Length of string: 100 cm Table 1-Effect of oscillation amplitude The bob’s Time to complete 5 cycles Mass of bob: 0.025 grams Initial bob angle: (degrees) Horizontal amplitude displacement Trial 1 -Sec Trail 2 -Sec Trial 3 -Sec Avg. 5 cycles. Std. Dev. 5 cycles Estimated period 1 cycle. 5 deg 32 cm 10 9.5 10 9.83 ,288 2.1 10 deg 52 cm 9.5 9.8 9.9 9.73 .208 1.9 15 deg 59 cm 9.9 10 9.7 9.866 .152 1.98 20 deg 70 cm 9.9 10 9.8 9.9 .1 1.98 25 deg 84 cm 9.9 10 9.9 9.933 .057 1.98 30 deg 97 cm 9.7 10.1 10 9.933 .208 1,94 Using the approximated formula for the period of the oscillation, and the measures of period (T) found in Table 1, estimate the value of g for each initial angle. Assuming a value of g equal to 9.81 m/s 2, calculate the relative error for each initial angle. Record your results in Table 1.a. formula: t=2pi sqrt l/g…

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