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Simple Harmonic Motion 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. 1 f 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? 15 s T 0.50 s 30 cylces x F Period: T = 0.500 s 1 1 f T 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 Hooke’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 The period and frequency of a wave • the period T of a wave is the amount of time it takes to go through 1 cycle • the frequency f is the number of cycles per second – the unit of a cycle-per-second is commonly referred to as a hertz (Hz), after Heinrich Hertz (1847-1894), who discovered radio waves. • frequency and period are related as follows: 1 f T • Since a cycle is 2p radians, the relationship between frequency and angular frequency is: 2pf t T The Reference Circle The reference circle compares the circular motion of an object with its horizontal projection. x A cos t x A cos(2p ft ) x = Horizontal displacement. A = Amplitude (xmax). = Reference angle. 2pf Velocity in SHM The velocity (v) of an oscillating body at any instant is the horizontal component of its tangential velocity (vT). vT = R = A; 2pf v = -vT sin ; = t v = - A sin t v = -2pf A sin 2pf 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 = -ac cos(t) v2 2 R2 ac ; ac 2 R R R a = -2A cos(t) R=A a 4p 2 f 2 A cos(2p ft ) a 4p f x 2 2 The Period and Frequency as a Function of a and x. For any body undergoing simple harmonic motion: Since a = -4p2f2x 1 f 2p a x and T = 1/f x T 2p 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. Springs and SHM • Attach an object of mass m to the end of a spring, pull it out to a distance A, and let it go from rest. The object will then undergo simple harmonic motion: x (t ) A cos(t ) v (t ) A sin( t ) a(t ) A 2 cos(t ) – Use Newton’s 2nd law, together with Hooke’s law, and the above description of the acceleration to find: k m 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: 1 f 2p k m m T 2p 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 = -0.2 m 1 f 2p x=0 k 1 m 2p f = 2.25 Hz x = +0.2 m 400 N/m 2 kg 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 = A a 4p f x 4p (2.25 Hz) (0.2 m) 2 2 2 a = 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 = -2pf A sin 2pf t x = -0.2 m x = 0 x = +0.2 m v 2p (2.25 Hz)(0.2 m)sin 2p (2.25 Hz)(2.69 s) (Note: in rads) v 2p (2.25 Hz)(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 v m x A cos(2p ft ) x = -0.2 m x = 0 x 0.12 m cos(2p ft ) ; A 0.20 m 2p ft 2.214 rad; x = +0.2 m (2p ft ) cos 1 ( 0.60) 2.214 rad t 2p (2.25 Hz) t = 0.157 s To Be Continued… Work Done in Stretching a Spring Work done ON the spring is positive; work BY spring is negative. x From Hooke’s law the force F is: F (x) = kx F F m To stretch spring from x1 to x2 , work is: Work ½kx ½kx 2 2 x1 x2 2 1 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 Hooke’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: Work ½kx ½kx 2 2 8 cm 0 m 2 1 U ½kx ½(196 N/m)(0.08 m) 2 U = 0.627 J F 2 Displacement in SHM x m x = -A x=0 x = +A • Displacement is positive when the position is to the right of the equilibrium position (x = 0) and negative when located to the left. • The maximum displacement is called the amplitude A. Velocity in SHM v (-) v (+) m x = -A x=0 x = +A • Velocity is positive when moving to the right and negative when moving to the left. • 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 • Acceleration is in the direction of the restoring force. (a is positive when x is negative, and negative when x is positive.) F ma kx • 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 ma kx or kx a 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? kx a m (400 N/m)(+0.07 m) a 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 xmax = A kA 400 N( 0.12 m) a m 2 kg Maximum Acceleration: m amax = ± 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: ½mvA2 + ½kxA 2 = ½mvB2 + ½kxB 2 Energy of a Vibrating System: A x a v m x = -A x=0 B x = +A • At points A and B, the velocity is zero and the acceleration is a maximum. The total energy is: U + K = ½kA2 x = A and v = 0. • At any other point: U + K = ½mv2 + ½kx2 Velocity as Function of Position. x a v m x = -A 1 2 x=0 k v A2 x 2 m mv kx kA 2 1 2 2 1 2 vmax when x = 0: x = +A 2 v k A m 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? ½mv2 + ½kx 2 = ½kA2 k v A2 x 2 m 800 N/m v (0.1 m) 2 (0.06 m) 2 2 kg v = ±1.60 m/s m +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 ½mv2 + ½kx 2 = ½kA2 v k 800 N/m A (0.1 m) m 2 kg v = ± 2.00 m/s m +x The Simple Pendulum The period of a simple pendulum is given by: L T 2p g L For small angles . 1 f 2p g 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 2p g 2 2 L T 4p ; g (2 s) 2 (9.8 m/s 2 ) L 2 4p L T 2g L= 2 4p L = 0.993 m To Be Continued…