{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework9-F08

Homework9-F08 - Homework 9 Chapt 14 solutions(1 Cosine Wave...

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

Homework 9 Chapt. 14 solutions (1) Cosine Wave The graph shows the position of an oscillating object as a function of time . The equation of the graph is , Part A What is A in the equation? Part B What is in the equation? Part C What is in the equation? From the figure we see when t = -N, x = M Using the cos wave function M = M cos(2 π /T * (-N) + Φ ) cos(2 π /T * (-N) + Φ )=1 (2 π /T * (-N) + Φ ) = 0, Φ = 2 π N /T

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

View Full Document
14.14. Model: The oscillating mass is in simple harmonic motion. Solve: (a) The amplitude A = 2.0 cm. (b) The period is calculated as follows: 2 2 10 rad s 0.628 s 10 rad s T T π π ω = = = = (c) The spring constant is calculated as follows: ( ) ( ) 2 2 0.050 kg 10 rad s 5.0 N m k k m m ω ω = = = = (d) The phase constant 1 0 4 rad. φ π = - (e) The initial conditions are obtained from the equations ( ) ( ) ( ) ( ) ( ) ( ) 1 1 4 4 2.0 cm cos 10 and 20.0 cm s sin 10 x x t t v t t π π = - = - - At t = 0 s, these equations become ( ) ( ) ( ) ( ) 1 1 0 0 4 4 2.0 cm cos 1.414 cm and 20.0 cm s sin 14.14 cm s x x v π π = - = = - - = In other words, the mass is at + 1.414 cm and moving to the right with a velocity of 14.14 cm/s. (f) The maximum speed is ( ) ( ) max 2.0 cm 10 rad s 20.0 cm s v A ω = = = . (g) The total energy ( )( ) 2 2 3 1 1 2 2 5.0 N m 0.02 m 1.0 10 J E kA - = = = × . (h) At t = 0.40 s, the velocity is ( ) ( )( ) 1 0 4 20.0 cm s sin 10 rad s 0.40 s 1.46 cm s x v π = - - = 14.19 Model: Assume a small angle of oscillation so there is simple harmonic motion.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}