Periodic Motion Simple periodic motion is that 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. Period, Period T, is the time for one complete oscillation. (seconds, s) Amplitude A Frequency, 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? x F Period: 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

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 x 1 x 2 m F To stretch spring from x 1 to x 2 , work is: (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/s 2) = 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: 0 U = 0. 627 J 8 cm F m

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. ) • Acceleration is a maximum at the end points and it is zero at the center of oscillation.

Acceleration vs. Displacement x a v m x = -A x=0 x = +A Given the spring constant, the displacement, and the mass, the acceleration can be found from: or 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 = -14. 0 m/s 2 a m Note: When the displacement is +7 cm (downward), the acceleration is -14. 0 m/s 2 (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 m Maximum Acceleration: amax = ± 24. 0 m/s 2 +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. x a v m x = -A x=0 x = +A For any two points A and B, we may write: ½mv. A 2 + ½kx. A 2 = ½mv. B 2 + ½kx. B 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 = ½k. A 2 x = A and v = 0. • At any other point: U + K = ½mv 2 + ½kx 2

Velocity as Function of Position. x a v m x = -A vmax when x = 0: x=0 x = +A

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? ½mv 2 + ½kx 2 = ½k. A 2 m v = ± 1. 60 m/s +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 ½mv 2 + ½kx 2 = ½k. A 2 m v = ± 2. 00 m/s +x

The Reference Circle The reference circle compares the circular motion of an object with its horizontal projection. x = Horizontal displacement. A = Amplitude (xmax). q = Reference angle. w = 2 f

Velocity in SHM The velocity (v) of an oscillating body at any instant is the horizontal component of its tangential velocity (v. T). v. T = w. R = w. A; w = 2 f v = -v. T sin ; = wt v = -w A sin w t v = -2 f A sin 2 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 q = -ac cos(wt) R=A a = -w 2 A cos(wt)

The Period and Frequency as a Function of a and x. For any body undergoing simple harmonic motion: Since a = -4 2 f 2 x and T = 1/f 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: -kx 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 x=0 f = 2. 25 Hz x = +0. 2 m

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) a v x m x = -0. 2 m x=0 x = +0. 2 m Acceleration is a maximum when x = A a = 40 m/s 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. ) a v x m v = -2 f A sin 2 f t x = -0. 2 m x = 0 x = +0. 2 m (Note: q in rads) 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 a v x m x = -0. 2 m x = 0 t = 0. 157 s x = +0. 2 m

The Simple Pendulum The period of a simple pendulum is given by: L For small angles q. 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 L = 0. 993 m

The Torsion Pendulum The period T of a torsion pendulum is given by: Where k’ 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 k’ is 0. 025 N m/rad. Find the period. (Neglect the torsion in the wire) For Disk: Disk I = ½m. R 2 I = ½(0. 16 kg)(0. 12 m)2 = 0. 00115 kg m 2 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

Summary (Cont. ) Hooke’s Law: In a spring, there is a restoring force that is proportional to the displacement. x The spring constant k is defined by: m F

Summary (SHM) x a v m x = -A x=0 x = +A Conservation of Energy: ½mv. A 2 + ½kx. A 2 = ½mv. B 2 + ½kx. B 2

Summary (SHM)

Summary: Period and Frequency for Vibrating Spring. x a v m x = -A x=0 x = +A

Summary: Simple Pendulum and Torsion Pendulum L