Chapter 13 Oscillations About Equilibrium

FOCUSED LEARNING TARGET GIVEN VIBRATIONS AND OSCILLATIONS CAUSED BY SPRING AND PENDULUM , I WILL BE ABLE TO CALCULATE FORCE OF SPRING (FS) , TOTAL ENERGY IN TERMS OF ELASTIC POTENTIAL ENERGY (US) AND KINETIC ENERGY (K), FREQUENCY (f) AND PERIOD (t) USING THE FOLLOWING EQUATIONS FS = -kx ; K= ½ mv2 ; Ki + Ui = Kf +Uf US = ½ kx2= ½ kA2 ; E = K + U ; T= 1/f ; f = 1/T f = 1/2π √k/m ; T = 2π√m/k ; vmax = A√k/m

HOMEWORK : CHAPTER 13. 1- 13.7 SUMMARIES 1 EXAMPLE FOR EACH SECTION 2 HW PROBLEMS(ODD ) FOR EACH SECTION 2 PROBLEMS IN THE COLLEGE BOARD https://apstudent.collegeboard.org/apcourse/ap-physics-b/exam-practice

Experiment :PENDULUM 2. HYPOTHESIS : ________________ 1. HYPOTHESIS : WHAT WILL HAPPEN TO THE NUMBER OF VIBRATIONS OF THE PENDULUM WHEN THE MASS OF THE PENDULUM INCREASES? ________________ 2. HYPOTHESIS : WHAT WILL HAPPEN TO THE NUMBER OF VIBRATIONS OF THE PENDULUM WHEN THE STRING OF THE PENDULUM INCREASES? ________________

CW :Pendulum Part 1:Half of the String’s Length 3. Length of the String _______cm 4. How many complete cycles in one minute ? Do 3 trials and get the average Cycles = _______ Time = 1 minute 5. Frequency in cycles per minute = ___________ 6.Frequency in cycles per second =_____________ 7. Period =_____sec

Pendulum Part 2 : Double the Length of the String 10. Frequency in cycles per minute = ___________ 11.Frequency in cycles per second =_____________ 12. Period = ________sec 8. Length of the String _______cm 9. How many complete cycles in one minute ? Do 3 trials and get the average Cycles = ______ Time = 1 minute

Pendulum Part 3 : Double the Mass and Use Half of the String’s Length 15. Frequency in cycles per minute = ___________ 16.Frequency in cycles per second =_____________ 17. Period = ____sec 13. Length of the String _______cm 14. How many complete cycles in one minute ? Do 3 trials and get the average Cycles = ______ Time = 1 minute

Conclusion 15. WHAT HAPPENS TO THE NUMBER OF VIBRATIONS OF THE PENDULUM WHEN THE STRING OF THE PENDULUM INCREASES? 16. WHAT HAPPENS TO THE NUMBER OF VIBRATIONS OF THE PENDULUM WHEN THE MASS OF THE PENDULUM INCREASES?

Read p 444-446 Vocabulary : Oscillation Propagation Wave pulse Hooke’s Law Simple Harmonic Motion Displacement Amplitude 8. Period 9. Frequency 10. Hertz Equations Necessary :

1. Periodic Motion Motion that repeats itself over a fixed and reproducible period of time. The revolution of a planet about its sun is an example of periodic motion. The highly reproducible period (T) of a planet is also called its year.

2. Mechanical devices on earth can be designed to have periodic motion 2. Mechanical devices on earth can be designed to have periodic motion. These devices are useful timers. They are called oscillators.

3. Simple Harmonic Motion You attach a weight to a spring, stretch the spring past its equilibrium point and release it. The weight bobs up and down with a reproducible period, T. Plot position vs. time to get a graph that resembles a sine or cosine function. The graph is "sinusoidal", so the motion is referred to as simple harmonic motion. Springs and pendulums undergo simple harmonic motion and are referred to as simple harmonic oscillators.

4. Crest 5. Troughs 6. Amplitude- Maximum displacement from equilibrium. Related to energy. 7. Equilibrium

7. Period(T) Length of time required for one oscillation.

8. Frequency How fast the oscillator is oscillating. f = 1/T Unit: Hz or s-1

9. Springs Springs are a common type of simple harmonic oscillator. Our springs are "ideal springs", which means • They are massless. • They are both compressible and extensible. They will follow Hooke's Law. • Fs = -kx

11. F=-kx ( Fs opposite with X)

Fs = -kx The acceleration of the block is equal to a = Fs / m Another way to describe the block’s motion is the energy it transfers. Us = ½ k x2 When you pull the block, you are increasing the elastic potential energy.

14. Releasing the block , potential energy becomes kinetic energy as the block moves. As it passes through equilibrium Us =0 , so all energy is K. 15. As it passes again through equilibrium , it compresses the spring , K –kinetic becomes Us- elastic potential

11. F=-kx ( Fs opposite with X) Us = 0 X > 0 K is maximized FS Us =maximized K =0 V=0 Us =maximized K =0 V=0 X< 0

13. A 12 cm long spring has a force constant (k) of 400 N/m 13. A 12 cm long spring has a force constant (k) of 400 N/m . How much force is required to stretch the spring to a length of 14cm.

13. A 12 cm long spring has a force constant (k) of 400 N/m 13. A 12 cm long spring has a force constant (k) of 400 N/m . How much force is required to stretch the spring to a length of 14cm. F = -kx F = - 400N/m ( .14m -.12m) = - 8 N

Conservation of Energy Springs and pendulums obey conservation of energy. The equilibrium position has high kinetic energy and low potential energy. The positions of maximum displacement have high potential energy and low kinetic energy. Total energy of the oscillating system is constant.

14. A block of mass m = 2 kg is attached to an ideal spring of force constant k = 500N/m . The amplitude of the resulting oscillations is 8 cm . Determine the total energy of the oscillator and the speed of the block when it is 4 cm from equilibrium.

E = Us + K = ½ kx2 + 0 = ½ (500N/m)(.08m)2 =1.6J v = 1.1 m/s 14. A block of mass m = 2 kg is attached to an ideal spring of force constant k = 500N/m . The amplitude of the resulting oscillations is 8 cm . Determine the total energy of the oscillator and the speed of the block when it is 4 cm from equilibrium. E = Us + K = ½ kx2 + 0 = ½ (500N/m)(.08m)2 =1.6J 1.6J= ½ kx2 + ½ mv2 = ½ (500N/m)(.04m)2 + ½ (2kg) v2 v = 1.1 m/s

14. A block of mass m = 0.05 kg oscillates on a spring whose force constant k is 500 N/m. The amplitude of the oscillations is 4 cm . Calculate the maximum speed of the block .

14. A block of mass m = 0.05 kg oscillates on a spring whose force constant k is 500 N/m. The amplitude of the oscillations is 4 cm . Calculate the maximum speed of the block . Us = ½ kx2  K = ½ mv2 ½ kx2 = ½ mv2 v = √ kX2 /m v =√ 500N/m ( .04m)2/ 0.05kg v = 4m/s

15. A block of mass m = 8kg is attached to an ideal spring of force constant k = 500N/m . The block is at rest at its equilibrium position. An impulsive force acts on a block , giving it an initial speed of 2m/s . Find the amplitude of the resulting oscillations?

15. A block of mass m = 8kg is attached to an ideal spring of force constant k = 500N/m . The block is at rest at its equilibrium position. An impulsive force acts on a block , giving it an initial speed of 2m/s . Find the amplitude of the resulting oscillations? Ei = Ef Ki + Ui = Kf + Uf ½ mv2 + 0 = 0 + ½ kx2 8kg (2m/s)2= 500N/m X2 x = 0.25 m

CW : A mass of 0.5 kg is connected to a massless spring with a force constant k of 50N/m . The system is oscillating on a frictionless horizontal surface . If the amplitude of the oscillations is 2cm , what is the total energy of the system ?

16. A block oscillating on the end of a spring moves from its position of a maximum spring stretch to maximum spring compression in 0.25sec . Determine the period and frequency of this motion.

16. A block oscillating on the end of a spring moves from its position of a maximum spring stretch to maximum spring compression in 0.25sec . Determine the period and frequency of this motion.

16. A block oscillating on the end of a spring moves from its position of a maximum spring stretch to maximum spring compression in 0.25sec . Determine the period and frequency of this motion. For whole cycle T = 0.5sec f = 1/T = 1/0.5s = 2 Hertz

CW: . A student observing an oscillating block counts 45.5 cycles of oscillations in one minute . Determine its frequency in hertz and period in seconds.

CW . A student observing an oscillating block counts 45.5 cycles of oscillations in one minute . Determine its frequency in hertz and period in seconds. f= 45.5 cycles/min X 1min/60sec = 0.758 cycles/sec = 0.758 Hz T = 1/f = 1/ 0.758Hz= 1.32 sec

17. A block of mass m = 2 kg is attached to a spring whose force constant k , is 300 N/m . Calculate the frequency and period of the oscillations of this spring –block system.

17. A block of mass m = 2 kg is attached to a spring whose force constant k , is 300 N/m . Calculate the frequency and period of the oscillations of this spring –block system. f = 1/2π √k/m f = 1/2π√ (300N/m) / 2kg f = 1.9 Hz T = 1/f = 1/ 1.9Hz = 0.51 sec

18. A block is attached to a spring and set into oscillatory motion and its frequency is measured . If this block were removed and replaced by a second block with ¼ the mass of the first block , how would the frequency of the oscillations compare to the first block ?

18. A block is attached to a spring and set into oscillatory motion and its frequency is measured . If this block were removed and replaced by a second block with ¼ the mass of the first block , how would the frequency of the oscillations compare to the first block ? f = 1/2π √ k/m = 1/2π √ k / (1/4)m f = 1/2π√ 4k/m = 1/2π (2) √k/m f increased by a factor of 2

CW: Calculate the period of a 300-g mass attached to an ideal spring with a force constant of 25 N/m.

CW A 300-g mass attached to a spring undergoes simple harmonic motion with a frequency of 25 Hz. What is the force constant of the spring?

CW : An 80-g mass attached to a spring hung vertically causes it to stretch 30 cm from its unstretched position. If the mass is set into oscillation on the end of the spring, what will be the period?

Sample Problem You wish to double the force constant of a spring. You A. Double its length by connecting it to another one just like it. B. Cut it in half. C. Add twice as much mass. D. Take half of the mass off.

Sample Problem You wish to double the force constant of a spring. You A. Double its length by connecting it to another one just like it. B. Cut it in half. C. Add twice as much mass. D. Take half of the mass off.

CW : Sample problem. A 2.0-kg mass attached to a spring oscillates with an amplitude of 12.0 cm and a frequency of 3.0 Hz. What is its total energy?

19. Pendulums A simple pendulum consists of a weight of mass m attached to a string or a massless rod that swings without friction, about the vertical equilibrium position . The pendulum can be thought of as a simple harmonic oscillator. The displacement needs to be small for it to work properly.

21. RESTORING FORCE FRESTORING = mg Sinθ The restoring force is provided by gravity. Displacement is zero at equilibrium. At the endpoints of the oscillation region , the restoring force and tangential acceleration at have the greatest magnitudes, the speed of the pendulum is zero , potential energy is maximized. As the pendulum passes through the equilibrium position, its kinetic energyand speed are maximized.

22. A simple pendulum has a period of 1s on earth 22. A simple pendulum has a period of 1s on earth. What would be its period on the Moon( where g = 1/6 of the earth ) T= 2π√ L/g T = increased by √6 = 1sec X √6

Sample problem Predict the period of a pendulum consisting of a 500 gram mass attached to a 2.5-m long string.

Sample problem Suppose you notice that a 5-kg weight tied to a string swings back and forth 5 times in 20 seconds. How long is the string?

Sample problem The period of a pendulum is observed to be T Sample problem The period of a pendulum is observed to be T. Suppose you want to make the period 2T. What do you do to the pendulum?

Conservation of Energy Pendulums also obey conservation of energy. The equilibrium position has high kinetic energy and low potential energy. The positions of maximum displacement have high potential energy and low kinetic energy. Total energy of the oscillating system is constant.