Chapter 14 - Simple Harmonic Motion A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University В© 2007 Photo by Mark Tippens A TRAMPOLINE exerts a restoring force on the jumper that is directly proportional to the average force required to displace the mat. Such restoring forces provide the driving forces necessary for objects that oscillate with simple harmonic motion. Objectives: After finishing this unit, you should be able to: вЂў Write and apply HookeвЂ™s Law for objects moving with simple harmonic motion. вЂў Write and apply formulas for finding the frequency f, period T, velocity v, or acceleration a in terms of displacement x or time t. вЂў Describe the motion of pendulums and calculate the length required to produce a given frequency. Periodic Motion Simple periodic motion is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a definite interval of time. f пЂЅ 1 T Amplitude A Period, T, is the time for one complete oscillation. (seconds,s) Frequency, f, is the number of complete oscillations per second. Hertz (s-1) Example 1: The suspended mass makes 30 complete oscillations in 15 s. What is the period and frequency of the motion? T пЂЅ 15 s пЂЅ 0.50 s 30 cylces x F Period: T = 0.500 s f пЂЅ 1 T пЂЅ 1 0.500 s Frequency: f = 2.00 Hz Simple Harmonic Motion, SHM Simple harmonic motion is periodic motion in the absence of friction and produced by a restoring force that is directly proportional to the displacement and oppositely directed. x F A restoring force, F, acts in the direction opposite the displacement of the oscillating body. F = -kx HookeвЂ™s Law When a spring is stretched, there is a restoring force that is proportional to the displacement. F = -kx x m F The spring constant k is a property of the spring given by: k= DF Dx Work Done in Stretching a Spring Work done ON the spring is positive; work BY spring is negative. x From HookeвЂ™s law the force F is: F (x) = kx F F m To stretch spring from x1 to x2 , work is: W ork пЂЅ ВЅ kx пЂ ВЅ kx 2 2 x1 x2 2 1 (Review module on work) Example 2: A 4-kg mass suspended from a spring produces a displacement of 20 cm. What is the spring constant? The stretching force is the weight (W = mg) of the 4-kg mass: 20 cm F = (4 kg)(9.8 m/s2) = 39.2 N F m Now, from HookeвЂ™s law, the force constant k of the spring is: k= DF Dx = 39.2 N 0.2 m k = 196 N/m Example 2(cont.: The mass m is now stretched a distance of 8 cm and held. What is the potential energy? (k = 196 N/m) The potential energy is equal to the work done in stretching the spring: 8 cm 0 m W ork пЂЅ ВЅ kx пЂ ВЅ kx 2 2 2 1 U пЂЅ ВЅ kx пЂЅ ВЅ (196 N /m)(0.08 m) 2 U = 0.627 J F 2 Displacement in SHM x m x = -A x=0 x = +A вЂў Displacement is positive when the position is to the right of the equilibrium position (x = 0) and negative when located to the left. вЂў The maximum displacement is called the amplitude A. Velocity in SHM v (-) v (+) m x = -A x=0 x = +A вЂў Velocity is positive when moving to the right and negative when moving to the left. вЂў It is zero at the end points and a maximum at the midpoint in either direction (+ or -). Acceleration in SHM +a -x +x -a m x = -A x=0 x = +A вЂў Acceleration is in the direction of the restoring force. (a is positive when x is negative, and negative when x is positive.) F пЂЅ m a пЂЅ пЂ kx вЂў Acceleration is a maximum at the end points and it is zero at the center of oscillation. Acceleration vs. Displacement a v x m x = -A x=0 x = +A Given the spring constant, the displacement, and the mass, the acceleration can be found from: F пЂЅ m a пЂЅ пЂ kx or a пЂЅ пЂ kx m Note: Acceleration is always opposite to displacement. Example 3: A 2-kg mass hangs at the end of a spring whose constant is k = 400 N/m. The mass is displaced a distance of 12 cm and released. What is the acceleration at the instant the displacement is x = +7 cm? a пЂЅ пЂ kx m aпЂЅ пЂ (400 N /m )(+0.07 m ) 2 kg a = -14.0 m/s2 a m Note: When the displacement is +7 cm (downward), the acceleration is -14.0 m/s2 (upward) independent of motion direction. +x Example 4: What is the maximum acceleration for the 2-kg mass in the previous problem? (A = 12 cm, k = 400 N/m) The maximum acceleration occurs when the restoring force is a maximum; i.e., when the stretch or compression of the spring is largest. F = ma = -kx aпЂЅ пЂ kA пЂЅ xmax = п‚± A пЂ 400 N ( п‚± 0.12 m ) m Maximum Acceleration: m 2 kg amax = В± 24.0 m/s2 +x Conservation of Energy The total mechanical energy (U + K) of a vibrating system is constant; i.e., it is the same at any point in the oscillating path. a v x m x = -A x=0 x = +A For any two points A and B, we may write: ВЅmvA2 + ВЅkxA 2 = ВЅmvB2 + ВЅkxB 2 Energy of a Vibrating System: A x a v m x = -A x=0 B x = +A вЂў At points A and B, the velocity is zero and the acceleration is a maximum. The total energy is: U + K = ВЅkA2 x = п‚± A and v = 0. вЂў At any other point: U + K = ВЅmv2 + ВЅkx2 Velocity as Function of Position. x a v m x = -A 1 2 mv пЂ« 2 1 2 x=0 kx пЂЅ 2 1 2 kA vmax when x = 0: x = +A k vпЂЅ 2 m v пЂЅ k m A A пЂx 2 2 Example 5: A 2-kg mass hangs at the end of a spring whose constant is k = 800 N/m. The mass is displaced a distance of 10 cm and released. What is the velocity at the instant the displacement is x = +6 cm? ВЅmv2 + ВЅkx vпЂЅ k 2 = ВЅkA2 A пЂx 2 2 m vпЂЅ 800 N /m m (0.1 m ) пЂ (0.06 m ) 2 2 kg v = В±1.60 m/s 2 +x Example 5 (Cont.): What is the maximum velocity for the previous problem? (A = 10 cm, k = 800 N/m, m = 2 kg.) The velocity is maximum when x = 0: 0 ВЅmv2 + ВЅkx 2 = ВЅkA2 vпЂЅ k m AпЂЅ 800 N /m (0.1 m ) 2 kg v = В± 2.00 m/s m +x The Reference Circle The reference circle compares the circular motion of an object with its horizontal projection. x пЂЅ A cos пЃ± пЃ± пЂЅ wt x пЂЅ A cos(2пЃ° ft ) x = Horizontal displacement. A = Amplitude (xmax). пЃ± = Reference angle. w пЂЅ 2пЃ°f Velocity in SHM The velocity (v) of an oscillating body at any instant is the horizontal component of its tangential velocity (vT). vT = wR = wA; w пЂЅ 2пЃ°f v = -vT sin пЃ± ; пЃ± = wt v = -w A sin w t v = -2пЃ°f A sin 2пЃ°f t Acceleration Reference Circle The acceleration (a) of an oscillating body at any instant is the horizontal component of its centripetal acceleration (ac). a = -ac cos пЃ± = -ac cos(wt) ac пЂЅ v w R 2 2 пЂЅ R R R=A 2 ; ac пЂЅ w R 2 a = -w2A cos(wt) a пЂЅ пЂ 4пЃ° f A cos(2пЃ° ft ) 2 2 a пЂЅ пЂ 4пЃ° f x 2 2 The Period and Frequency as a Function of a and x. For any body undergoing simple harmonic motion: Since a = -4пЃ°2f2x f пЂЅ 1 пЂa 2пЃ° x and T пЂЅ 2пЃ° T = 1/f пЂx a The frequency and the period can be found if the displacement and acceleration are known. Note that the signs of a and x will always be opposite. Period and Frequency as a Function of Mass and Spring Constant. For a vibrating body with an elastic restoring force: Recall that F = ma = -kx: f пЂЅ 1 k 2пЃ° m T пЂЅ 2пЃ° m k The frequency f and the period T can be found if the spring constant k and mass m of the vibrating body are known. Use consistent SI units. Example 6: The frictionless system shown below has a 2-kg mass attached to a spring (k = 400 N/m). The mass is displaced a distance of 20 cm to the right and released. What is the frequency of the motion? x a v m x=0 x = -0.2 m f пЂЅ 1 k 2пЃ° m пЂЅ x = +0.2 m 1 400 N /m 2пЃ° 2 kg f = 2.25 Hz Example 6 (Cont.): Suppose the 2-kg mass of the previous problem is displaced 20 cm and released (k = 400 N/m). What is the maximum acceleration? (f = 2.25 Hz) x a v m x=0 x = -0.2 m x = +0.2 m Acceleration is a maximum when x = п‚± A a пЂЅ пЂ 4пЃ° f x пЂЅ пЂ 4пЃ° (2.25 H z) ( п‚± 0.2 m) 2 2 2 a = п‚± 40 m/s2 2 Example 6: The 2-kg mass of the previous example is displaced initially at x = 20 cm and released. What is the velocity 2.69 s after release? (Recall that f = 2.25 Hz.) x a v m v = -2пЃ°f A sin 2пЃ°f t x = -0.2 m x = 0 x = +0.2 m v пЂЅ пЂ 2пЃ° (2 .2 5 H z)(0 .2 m ) sin пЃ› 2 пЃ° (2 .2 5 H z)(2 .6 9 s) пЃќ (Note: пЃ± in rads) v пЂЅ пЂ 2пЃ° (2.25 H z)(0.2 m )(0.324) v = -0.916 m/s The minus sign means it is moving to the left. Example 7: At what time will the 2-kg mass be located 12 cm to the left of x = 0? (A = 20 cm, f = 2.25 Hz) -0.12 m x a m x пЂЅ A cos(2пЃ° ft ) x = -0.2 m x = 0 cos(2пЃ° ft ) пЂЅ x A пЂЅ v пЂ 0.12 m x = +0.2 m пЂ1 (2 пЃ° ft ) пЂЅ cos ( пЂ 0.60) ; 0.20 m 2пЃ° ft пЂЅ 2.214 rad; tпЂЅ t = 0.157 s 2.214 rad 2 пЃ° (2.25 H z) The Simple Pendulum The period of a simple pendulum is given by: T пЂЅ 2пЃ° L L g For small angles пЃ±. f пЂЅ 1 g 2пЃ° L mg Example 8. What must be the length of a simple pendulum for a clock which has a period of two seconds (tick-tock)? L T пЂЅ 2пЃ° L g T 2 пЂЅ 4пЃ° 2 L 2 ; L = g 2 LпЂЅ T g 4пЃ° 2 2 (2 s) (9.8 m /s ) 4пЃ° 2 L = 0.993 m The Torsion Pendulum The period T of a torsion pendulum is given by: T пЂЅ 2пЃ° I k' Where kвЂ™ is a torsion constant that depends on the material from which the rod is made; I is the rotational inertia of the vibrating system. Example 9: A 160 g solid disk is attached to the end of a wire, then twisted at 0.8 rad and released. The torsion constant kвЂ™ is 0.025 N m/rad. Find the period. (Neglect the torsion in the wire) For Disk: I = ВЅmR2 I = ВЅ(0.16 kg)(0.12 m)2 = 0.00115 kg m2 T пЂЅ 2пЃ° I k' пЂЅ 2пЃ° 0.00115 kg m 2 0.025 N m /rad T = 1.35 s Note: Period is independent of angular displacement. Summary Simple harmonic motion (SHM) is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a definite interval of time. The frequency (rev/s) is the reciprocal of the period (time for one revolution). x m F f пЂЅ 1 T Summary (Cont.) HookeвЂ™s Law: In a spring, there is a restoring force that is proportional to the displacement. F пЂЅ пЂ kx x The spring constant k is defined by: m F k пЂЅ DF Dx Summary (SHM) x a v m x = -A x=0 F пЂЅ m a пЂЅ пЂ kx x = +A a пЂЅ пЂ kx m Conservation of Energy: ВЅmvA2 + ВЅkxA 2 = ВЅmvB2 + ВЅkxB 2 Summary (SHM) 1 2 vпЂЅ mv пЂ« 2 k 1 2 A пЂx 2 m x пЂЅ A cos(2пЃ° ft ) kx пЂЅ 2 2 1 2 kA v0 пЂЅ 2 k A m a пЂЅ пЂ 4пЃ° f x 2 v пЂЅ пЂ 2пЃ° fA sin(2пЃ° ft ) 2 Summary: Period and Frequency for Vibrating Spring. a v x m x = -A f пЂЅ f пЂЅ x=0 1 пЂa 2пЃ° x 1 k 2пЃ° m T пЂЅ 2пЃ° x = +A пЂx a T пЂЅ 2пЃ° m k Summary: Simple Pendulum and Torsion Pendulum f пЂЅ 1 g 2пЃ° L T пЂЅ 2пЃ° L T пЂЅ 2пЃ° L g I k' CONCLUSION: Chapter 14 Simple Harmonic Motion

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