{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# chap 13 - OSCILLATORY MOTION 13 1 1 = = 2.27 10 3 s f 440...

This preview shows pages 1–3. Sign up to view the full content.

13.1 OSCILLATORY MOTION EXERCISES Section 13.1 Describing Oscillatory Motion 16. T = 1/ f = 1/440 Hz = 2.27 ms (Equation 13.1). 17. INTERPRET The question here is about the oscillatory behavior of the violin string. Given the frequency of oscillation, we are asked to find the period. DEVELOP The relationship between period and frequency is given by Equation 13.1, T f = 1 . EVALUATE Using Equation 13.1, we obtain T f == = × 1 1 440 227 10 3 Hz s . ASSESS The period is the oscillation is the inverse of the frequency. Note that the unit of frequency is the hertz: 11 1 Hz s = . 18. Tf × = × = 1 1 8 66 10 1 15 10 1 5 13 14 // ( . . . Hz) s 1 fs (Equation 13.1). 19. INTERPRET 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 t = 0. DEVELOP The general expression of the position of an object undergoing simple harmonic motion is given by Equation 13.8: xt A t () cos ( ) =+ ωφ where A is the amplitude, ω is the angular frequency, and φ is the phase constant. By taking the time derivative of , we obtain the corresponding velocity as a function of time (see Equation 13.9): vt dx t dt At sin( ) + ωωφ EVALUATE (a) Since the displacement is a maximum at t = 0, the phase constant is zero: = 0. Use Equation 13.8 with A === 10 2 5 10 1 cm, Hz s ωπ π , the displacement is found to be A t t ( ) ( )cos [ ( )] = 10 10 1 cm s (b) Equation 13.9 shows that the maximum (positive) velocity occurs at t = 0 if sin =− 1 or φπ /. 2 Therefore, with A 25 5 1 ., , s we have A t t ( ) cos( ) ( . )cos[( ) ] ( = = 5 2 1 cm s / 5 1 . )sin[( ) ] cm s t where we have used cos( / ) sin . tt −= 2 ASSESS 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. INTERPRET From the number of oscillations in a given time, we calculate the frequency and the period of the oscillations. DEVELOP Use the definitions of period and of frequency: T is the time for a single oscillation, and f T = 1 . There are 9 oscillations in 60 seconds. EVALUATE T 60 667 s 9 cycles s. . f T 1 015 .H z . ASSESS This is a relatively slow oscillation, as you might expect for a building-sized object. 13

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
13.2 Chapter 13 Section 13.2 Simple Harmonic Motion 21. INTERPRET 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. DEVELOP Given the spring constant k and the mass, the frequency and period of the oscillation may be obtained from Equations 13.8b and c: f k m T f m k == = 1 2 1 2 π Using Equations 13.9 and 13.10, the maximum speed and the maximum force are: vA Fm am A max max max = ω 2 EVALUATE (a) For a mass on a spring, Equation 13.7a gives = k m 56 02 529 1 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 18

chap 13 - OSCILLATORY MOTION 13 1 1 = = 2.27 10 3 s f 440...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online