This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Physics 2214 Fall 2008 Solutions to Problem Set #1 1  1. For the object to undergo SHM when displaced by a small amount from a certain position x , the graph of F(x) vs. x must pass through zero at that point, with a negative slope. Points B and G satisfy these criteria. Explanation: The force must be zero at the equilibrium position. In addition, when the object is displaced in the positive x direction, the force must be negative (i.e. pointing in the negative x direction), to pull it back to its equilibrium position. Conversely, when the object is displaced in the negative x direction, the force must be positive, to pull it back to its equilibrium position. 2. (a) First consider the motion of Object 1. The equilibrium displacement, x eq1 , is halfway between the two vertical extremes (maxima and minima) of the (blue) sinusoid: x eq1 = 3.0 mm . The amplitude, A 1 , is the vertical distance from the equilibrium value of x to the maximum value of the sinusoid: A 1 = 2.5 mm . The angular frequency, ω 1 , is found from the period T 1 of the sinusoid: ω 1 = 2 π T 1 , where T 1 is the horizontal distance between adjacent peaks: T 1 = 0.80 s. Thus we have ω 1 = 2 π T 1 = 2 π 0.8 s = 2.5 π rad/s . The phase shift, φ 01 , is found from the time offset t 01 of the peak which is closest to the left of the t=0 axis, with respect to the t=0 axis: φ 01 = 2 π ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ t 01 T 1 . Reading from the graph: t 01 = 0.5 s. Then φ 01 = 2 π ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ t 01 T 1 = 2 π ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ 0.5 s 0.8 s = 1.25 π rad . The quantities x eq1 , A 1 , ω 1 , T 1 , and t 01 are all shown on the graph below.  2  Similarly, we can obtain the corresponding constants for Object 2 by measuring from the graph: x eq2 = 1.5 mm; A 2 = 1.5 mm; T 2 = 0.80 s; ω 2 = 2.5 π rad/s; t 02 = +0.15 s ; φ 02 = +0.375 π rad These quantities are also indicated on the above graph. (b) The equation of motion for Object 1 is x 1 (t) = x eq1 + A 1 cos ( ω t + φ 01 ) The instantaneous velocity v 1 (t) is the time derivative of x 1 (t): v 1 (t) = dx 1 dt = A 1 ω sin ( ω t + φ 01 ) ....
View
Full
Document
This note was uploaded on 12/24/2008 for the course PHYS 2214 taught by Professor Giambattista,a during the Fall '07 term at Cornell.
 Fall '07
 GIAMBATTISTA,A
 Physics

Click to edit the document details