Presentation is loading. Please wait.

Presentation is loading. Please wait.

Oscillatory Motion EXAMPLES

Similar presentations


Presentation on theme: "Oscillatory Motion EXAMPLES"— Presentation transcript:

1 Oscillatory Motion EXAMPLES
Chapter 15 Oscillatory Motion EXAMPLES

2 Chapter 15 Oscillatory Motion: EXAMPLES

3 Example 15-1: Motion of the Block (Horizontal)
The block continues to oscillate between –A and +A These are turning points of the motion The force F is conservative In the absence of friction, the motion will continue forever Real systems are generally subject to friction, so they do not actually oscillate forever

4 Example 15-1: Motion of the Block (Vertical)
When the block is hung from a vertical spring, its weight will cause the spring to stretch Figure (a): Free spring, hang vertically Figure (b): Mass m attached to spring in new equilibrium position, which occurs when: F = mg – kxo  xo = mg/k If the resting position of the spring is defined as x = 0, the same analysis as was done with the horizontal spring will apply to the vertical spring-mass system

5 Example 15-2: Conceptual Example (SHM Forces)
Which of the following represent an object undergoing SHM (a). F = – 0.5x2 (b). F = –2.3y (c). F = 8.6x (d). F = – 4θ Answer: (b). F = – 2.3y & (d). F = – 4θ Force has minus sign required to restore the system to equilibrium. Force is proportional to a displacement (which need not to be x).

6 15-3: SHM Example 1 Initial conditions at t = 0 : x (0)= A v (0) = 0
This means f = 0 The velocity reaches extremes of : ± ωA The acceleration reaches extremes of: ± ω2A

7 15-3: SHM Example 2 Initial conditions at t = 0 : x (0)=0 v (0) = vi
This means f = - π/2 The graph is shifted one-quarter cycle to the right compared to the graph of x (0) = A

8 Example 15-4: Conceptual Example (Doubling Amplitude)
Suppose the spring from figure is stretched twice as far (to x = 2A). What happened to: E, vmax, & amax? (a). E = ½kA2 (15.21) (b). vmax = ωA = (k/m)½A (15.17) (c). amax = ω2A = (k/m)A (15.18) Answers: (a). E = ½k(2A)2 = 4( kA2) So stretching A twice as far, quadruples the energy (b). vmax = ω(2A) = 2ωA So stretching A twice as far, doubles the maximum velocity (c). amax = ω2(2A) = 2ω2A So stretching A twice as far, doubles the maximum acceleration

9 Material from the book to Study!!!
Material for the Midterm Material from the book to Study!!! Objective Questions: 12-14 Conceptual Questions: 3-5 Problems:


Download ppt "Oscillatory Motion EXAMPLES"

Similar presentations


Ads by Google