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Unformatted text preview: OSCILLATORY MOTION E XERCISES Section 13.1 Describing Oscillatory Motion 16. T = 1/ f = 1/440 Hz = 2.27 ms (Equation 13.1). 17. I NTERPRET The question here is about the oscillatory behavior of the violin string. Given the frequency of oscillation, we are asked to find the period. D EVELOP The relationship between period and frequency is given by Equation 13.1, 1 . f T = E VALUATE Using Equation 13.1, we obtain 3 1 1 2.27 10 s 440 Hz T f = = = × A SSESS The period is the oscillation is the inverse of the frequency. Note that the unit of frequency is the hertz: 1 1 Hz 1s . = 18. 13 14 1/ 1/(8.66 10 Hz) 1.15 10 s 11.5 fs T f = = × = × = (Equation 13.1). 19. I NTERPRET The problem involves simple harmonic motion. We want to write down an expression that characterizes the oscillation, given its amplitude, frequency, and the function at 0. t = D EVELOP The general expression of the position of an object undergoing simple harmonic motion is given by Equation 13.8: ( ) cos( ) x t A t ϖ φ = + where A is the amplitude, ϖ is the angular frequency, and φ is the phase constant. By taking the time derivative of ( ), x t we obtain the corresponding velocity as a function of time (see Equation 13.9): ( ) ( ) sin( ) dx t v t A t dt ϖ ϖ φ = =  + E VALUATE (a) Since the displacement is a maximum at 0, t = the phase constant is zero: 0. φ = Use Equation 13.8 with 1 10 cm, 2 (5 Hz) 10 s , A ϖ π π = = = the displacement is found to be 1 ( ) cos( ) (10 cm)cos[(10 s ) ] x t A t t ϖ φ π = + = (b) Equation 13.9 shows that the maximum (positive) velocity occurs at t = if sin 1 φ =  or /2. φ π =  Therefore, with 1 2.5 cm, 5 s , A ϖ = = we have 1 1 ( ) cos( ) (2.5 cm)cos[(5 s ) /2] (2.5 cm)sin[(5 s ) ] x t A t t t ϖ φ π = + = = where we have used cos( /2) sin . t t ϖ π ϖ = A SSESS For a system undergoing simple harmonic motion, once the position as a function of time is given in the form of Equation 13.8, physical quantities such as velocity, acceleration, angular frequency and period can all be readily determined. 20. I NTERPRET From the number of oscillations in a given time, we calculate the frequency and the period of the oscillations. D EVELOP Use the definitions of period and of frequency: T is the time for a single oscillation, and 1 . T f = There are 9 oscillations in 60 seconds. E VALUATE 60 s 9 cycles 6.67 s. T = = 1 0.15 Hz. T f = = 13.1 13 13.2 Chapter 13 A SSESS This is a relatively slow oscillation, as you might expect for a buildingsized object. Section 13.2 Simple Harmonic Motion 21. I NTERPRET In this problem a mass attached to a spring undergoes simple harmonic motion. Given the mass, the spring constant, and the amplitude, we are asked to compute the frequency and the period of oscillation, the maximum velocity, and the maximum force in the spring....
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This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 Prof.Kim
 Physics

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