1. Simple Harmonic Motion Lesson 2

2. Work Done in Stretching a Spring
Work done ON the spring is positive; work BY spring is negative. F x m
From Hooke’s law the force F is:
F (x) = kx x1 x2 F
To stretch spring from x1 to x2 , work is:

3. Example 2: A 4-kg mass suspended from a spring produces a displacement of 20 cm. What is the spring constant? F
20 cm m
The stretching force is the weight (W = mg) of the 4-kg mass:
F = (4 kg)(9.8 m/s2) = 39.2 N
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

4. 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) F
8 cm m
The potential energy is equal to the work done in stretching the spring:
U = J

5. Displacement in SHM m
x = 0
x = +A
x = -A x
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.

6. Velocity in SHM m v (-) v (+)
x = 0
x = +A
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 -).

7. Acceleration in SHM m -x +x
x = +A
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.

8. Acceleration vs. Displacementx v a m
x = 0
x = +A
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.

9. 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? m +x a
a = m/s2
Note: When the displacement is +7 cm (downward), the acceleration is m/s2 (upward) independent of motion direction.

10. Maximum Acceleration:
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. m +x
F = ma = -kx
xmax =  A
Maximum Acceleration:
amax = ± 24.0 m/s2

11. 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 v a m
x = 0
x = +A
x = -A
For any two points A and B, we may write:
½mvA2 + ½kxA 2 = ½mvB2 + ½kxB 2

12. Energy of a Vibrating System:
x = 0
x = +A
x = -A x v a A B
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

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

14. 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? m +x
½mv2 + ½kx 2 = ½kA2
v = ±1.60 m/s

15. The velocity is maximum when x = 0:
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: m +x
½mv2 + ½kx 2 = ½kA2
v = ± 2.00 m/s

16. The Simple Pendulum L For small angles q. mg
The period of a simple pendulum is given by: mg L
For small angles q.

17. 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 = m

18. To Be Continued…