1. Oscillatory Motion
Serway & Jewett (Chapter 15)

2. M M M Equilibrium position: no net force
The spring force is always directed back towards equilibrium (hence called the ‘restoring force’). This leads to an oscillation of the block about the equilibrium position. M
For an ideal spring, the force is proportional to displacement. For this particular force behaviour, the oscillation is simple harmonic motion. x
F = -kx

3. SHM: x(t) A = amplitude t f = phase constant w = angular frequency
A is the maximum value of x (x ranges from +A to -A).
f gives the initial position at t=0: x(0) = A cosf .
w is related to the period T and the frequency f = 1/T
T (period) is the time for one complete cycle (seconds).
Frequency f (cycles per second or hertz, Hz) is the number of complete cycles per unit time.

4. units: radians/second or s-1
 (“omega”) is called the angular frequency of the oscillation.

5. Velocity and Acceleration
a(t) =- w 2 x(t)

6. Position, Velocity and Acceleration
x(t) t
v(t) t
a(t) t
Question: Where in the motion is the velocity largest?
Where in the motion is acceleration largest?

7. Example:
SHM can produce very large accelerations if the frequency is high.
Engine piston at 4000 rpm, amplitude 5 cm:

8. Simple Harmonic Motion
SHM:
We can differentiate x(t):
and find that acceleration is proportional to displacement:
a(t) = - w 2 x(t)
But, how do we know something will obey x=Acos(t) ???

9. Mass and Spring M Newton’s 2nd Law: F = -kx so x
This is a 2nd order differential equation for the function x(t). Recall that for SHM, a = -w 2 x : identical except for the proportionality constant. So the motion of the mass will be SHM:
x(t) = A cos (wt + f), and to make the equations for acceleration match, we require that
, or
(and w = 2p f, etc.).
Note: The frequency is independent of amplitude

10. Example: Elastic bands and a mass
Example: Elastic bands and a mass. A mass, m, is attached to two elastic bands. Each has tension T. The system is on a frictionless horizontal surface. Will this behave like a SHO?

11. Quiz
The ball oscillates vertically on a single spring with period T0 . If two identical springs are used, the new period will be
longer
shorter
by a factor of 2 4

12. Quiz
The ball oscillates vertically on a single spring with period T0 . If two identical springs are used, the period will be
longer
shorter
by a factor of 2 4 1

13. Quiz
μs=0.5 B
ω=10 s-1
The amplitude of the oscillation gradually increases till block B starts to slip. At what A does this happen? (there is no friction between the large block and the surface)
Any A
5 cm
50 cm
Not known without mB.

14. Solution

15. Look again at the block & spring
Energy in SHM M
Look again at the block & spring
We could also write E = K+U = ½ m(vmax )2

16. E
U, K oscillate back and forth “out of phase” with each other; the total E is constant.
n.b.! U, K go through two oscillations while the position x(t) goes through one. U K t T x t v

17. Suppose you double the amplitude of the motion:
1) What happens to the maximum speed?
Doubles
4 x Larger
Doesn’t change
2) What happens to the maximum acceleration?
Doubles
4 x Larger
Doesn’t change
3) What happens to the the total energy?
Doubles
4 x Larger
Doesn’t change

18. Summary
SHM:
(get v, a with calculus)
Definitions: amplitude, period, frequency, angular frequency, phase, phase constant.
The acceleration is proportional to displacement:
a(t) = -w2 x(t)
Practice problems, Chapter 15: 3, 5, 11, 19, 23 (6th ed – Chapter 13)
1, 3, 5, 9, 19, 23