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1 Chapter 7 Review Student Boo pages Knowledge. Oscillatory otion is otion that repeats itself at regular intervals. For exaple, a ass oscillating on a spring and a pendulu swinging bac and forth.. A ball will bounce with oscillatory otion if it has a unifor period. This eans that it ust bounce to the sae height in each oscillation. 3. An elastic aterial will defor due to a stress. Once the stress is reoved, the aterial will return to its original shape, without loss of energy. 4. The restoring force is the only force that acts on an isolated, frictionless, siple haronic oscillator. 5. The direction of the restoring force is always opposite to the displaceent. 6. The slope represents the spring constant. 7. It is not in its equilibriu position. x 8. Acceleration depends on displaceent, as given by a. Therefore, it is not unifor. 9. The restoring force varies with the sine of the angle of displaceent. For sall angles, this relationship is alost linear. As the displaceent angle increases, the relationship is no longer linear. This difference begins to show at approxiately If the forced frequency is siilar to the natural, resonant frequency, the aplitude of the oscillatory otion will be increased. The otion still taes place at the resonant frequency. Applications. (a) Analysis and Solution (a) (b) ote that the angles in the diagra have been exaggerated for illustration purposes. F g& Fgsin % g sin % ' ( #.0 g$ * 9.8 sin 5 + ), s

2 The restoring force on the pendulu is.5 [toward equilibriu]. (b) Analysis and Solution F g& Fgsin % g sin % ' ( #.0 g$ * 9.8 sin 5 + ), s The restoring force on the pendulu is 0.85 [toward equilibriu].. Analysis and Solution f T 0.00s.5 0 Hz. The frequency of the guitar string is.5. 0 Hz. 3. Analysis and Solution T f 0.67 Hz.5 s The period is.5 s. 4. Analysis and Solution f T 0.00 s 0.0 Hz The frequency is 0.0 Hz. 5. Analysis and Solution 34

3 F g& Fgsin % g sin % ' ( #.0 g$ * # sin 5.0) $, s - 5. The restoring force acting on the pendulu is 5. [toward equilibriu]. 6. Analysis and Solution Any two points fro the line of best fit can be used to deterine the slope. The following points were used for the solution below: (x, y ) = (0.0, 0.0) and (x, y ) = (0.75, 50) slope (50 0.0) ( ).0. 0 / The spring constant of the spring is.0. 0 /. 7. Given height of spring fro floor =.80 height of spring fro floor with 00.0-g ass attached =.30 Required (a) spring constant () (b) force needed to pull the 00.0-g ass through a displaceent of 0.0 c ( F ) (c) distance of the g ass ( x ) Analysis and Solution Draw diagra of the situation described in each part the proble. You need to now the spring constant of the spring to deterine the force and displaceents in parts (b) and (c). To deterine the spring constant, use the force of gravity for the 00.0-g ass. The displaceent of the ass-spring is 50.0 c down. 35

4 (a) x [down] Fg x Fg x g x ' ( # g$ * 9.8 s /.96 / (b) Tae the displaceent of the 00.0-g ass (.30 ) fro the floor as the new equilibriu position, and deterine the force required to pull it down 0.0 c. x 0.00 [down] F x (.96 )( 0.00 ) 0.39 (c) Deterine the distance for the g ass fro the spring s original equilibriu position (.80 ) fro the floor. Use the spring constant found in part (a) to deterine this distance. Fg x g ' ( ( g) * 9.8 s The height fro the floor is: To three significant digits, the distance is Paraphrase (a) The spring constant of the spring is.96 /. (b) A person ust apply a force of 0.39 to pull the hanging 00.0-g ass down through a displaceent of 0.0 c. (c) If the 00.0-g ass is reoved and a g ass is attached, it will hang fro the floor. 36

5 8. Given A 00.0 / B 50.0 / F 5.0 [right] Required cobined stretch of the two springs (x T ) Analysis and Solution Each spring will be stretched differently by the force. To deterine the total stretch, calculate the stretch of each spring then add the up. Spring A: A 00.0 / FA xa FA xa A Spring B: B 50.0 / FB xb FB xb B x x / x T A B 0.50 /

6 9. Paraphrase The cobined stretch of the two springs is Force vs. Displaceent The graph shows that the elastic band does not obey Hooe s law because the graph is not linear. 0. Analysis and Solution l T 0 g T g l ' ( #.00s$ * g #.00 s $ ' ( * 9.8 s or 4.8 c The length of the pendulu ust be 4.8 c.. Analysis and Solution T g s The period of the ass is 3.00 s. 38

7 . Analysis and Solution vax A g 0.0 /s The axiu speed of the crate is 0.0 /s. 3. Given A = 0.0 v ax = 0.00 /s = g = g x = [upward] Required acceleration ( a ) Analysis and Solution To deterine the acceleration of the ass, find the spring constant of its thread. Do this first and then use the spring constant in the acceleration equation. Mae sure to use the appropriate SI units for the ass. Tae up as positive, and down as negative. 39

8 v v A A vax A ax ax ( g) ' ( * +. s (0.0 ) /. Use this spring constant to deterine the acceleration of the ass at a displaceent of [up]. F x a x x a ' 3 ( * (0.000 ), g 0.3 /g 0.3 /s Paraphrase The ass experiences an acceleration of displaceent of [up]. 4. Analysis and Solution l T 0 g ' ( * 9.8 s +.0 s The period of the pendulu is.0 s. 0.3 /s [down], when it is at a

9 5. Analysis and Solution T f 0.8 Hz 5.49 s l g T # 5.49 s$ 0.65 /s 0.65 /g Pluto s gravitational field strength is 0.65 /g. 6. (a) X represents l g. (b) Analysis and Solution l X g l T g (.79 s) s l s ( ) g ' ( s 9.8 * s The length of the pendulu is Extensions 7. 4

10 (a) Analysis and Solution vax A g.6 /s The axiu speed of the ass is.6 /s. (b) Analysis and Solution T g s The ass will have a period of oscillation of s. The period is independent of the aplitude. 8. (a) Given f = Hz Required frequency of the oscillator when the ass is doubled (f) Analysis and Solution To find the frequency when the ass is doubled, odify the equation for the period to solve for frequency. Substitute in place of in the equation to deterine the factor by which the frequency increases or decreases. Then deterine the actual change to the frequency by ultiplying this factor by the original frequency. T 0 f T 0 0 4

11 Change to. f (.44) ' ( * +,.44-0 (0.707) 0 The frequency has changed by a factor of ow deterine the new frequency. f Hz = Hz Paraphrase The frequency of the oscillator will change fro Hz to Hz. (b) For any siple haronic oscillator, the aplitude of oscillation does not affect the period, and therefore does not affect the frequency. The frequency will stay at Hz. 9. (a) A bouncing ball is not a siple haronic oscillator because the force does not vary with displaceent. The force of gravity is constant regardless of the position of the ball. (b) The oveent of a puc on the ice cannot be considered SHM. The force applied by the stic is constant throughout its displaceent. (c) A pluced guitar string is an exaple of SHM. The guitar string s restoring force varies directly with its displaceent. 43