1.  One way to understand SHM is to reconsider the circular motion of a particle and rotational kinematics (The Reference Circle)  The particle travels on a circle of radius r=A with the line from the center to the particle making an angle  with respect to the x-axis at some instant in time  Now, project this 2D motion onto a 1D axis

2. A  x A x=Ax=-Ax=0 From Chap. 8  x is the displacement for SHM, which includes the motion of a spring  SHM is also called sinusoidal motion  x max = A = amplitude of the motion (maximum)   is the angular frequency (speed) in rad/s. It remains constant during the motion.

3.  As we saw in Chap. 8,  and the period T are related  Define the frequency  As an example, the alternating current (AC) of electricity in the US has a frequency of 60 Hz  Now, lets consider the velocity for SHM, again using the Reference circle Units of 1/s = Hertz (Hz) Relates frequency and angular frequency

4.  A x vTvT v  Amplitude of the velocity is  Acceleration of SHM  acac a x

5.  The amplitude of the acceleration is  Summary of SHM x (A=1)  t (  rads)

6. v (A  =1) a (A  2 =1)  t (  rads)

7. Example: Problem 10.22 Given an amplitude of 0.500 m and a frequency of 2.00 Hz, determine (a) the displacement, (b) the velocity, and (c) the acceleration at time 0.0500 s. Solution: Given: A=0.500 m, f=2.00 Hz, t=0.0500 s. (a)

8. Frequency of Vibration  Apply Newton’s 2 nd Law to the spring-mass system (neglect friction and air resistance)

9. 0A k m x FsFs Consider x-direction only Substitute x and a for SHM Angular frequency of vibration for a spring with spring constant k and attached mass m. Spring is assumed to be massless.

10.  This last equation can be used to determine k by measuring T and m Frequency of vibration Period of vibration Work and Potential Energy of a Spring  We know that work equals (force x displacement)

11. x0x0 xfxf x=0 But the force is not constant Take the average force Then the work done by the spring is