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SIMPLE HARMONIC MOTION

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Presentation on theme: "SIMPLE HARMONIC MOTION"— Presentation transcript:

1 SIMPLE HARMONIC MOTION

2 Simple harmonic motion- Any motion that is defined by period and amplitude
Period- The time for one cycle T= time/ #cycles Frequency- cycles per second. f = #cycles/ time f=1/T (inverse of the period)

3 Amplitude- Displacement from equilibrium.
Periodic Motion- The acceleration is directly proportional to the displacement of the objects equilibrium position. (vibrations, springs, pendulums and waves)

4 PENDULUM MOTION

5 Pendulum- Any object that swings back and forth about an axis
T = time/# of swings

6 Period of a pendulum is:
Independent of mass Independent of release angle Independent of weight Dependent only on the length of the pendulum.

7 *SORRY FOLKS- NOT ON THE
FORMULA T= 2π√ L/g T-period(sec) L-length (m) g= 9.81m/s2 *SORRY FOLKS- NOT ON THE REFERENCE TABLE*

8 Ex: What is the period of a pendulum that is 50cm long?
T= 2π√ L/g T= 2(3.14)√ 0.50m/9.81m/s2 T= 1.4sec

9 Graphs period period mass angle period length

10 KE and PE As the pendulum swings from its peak to the center, the potential energy is converted to kinetic energy.

11 GRAPHING THE SWING Displacement of the swing is equal to its amplitude
time Displacement of the swing is equal to its amplitude

12 MECHANICAL RESONANCE An increase in the amplitude
(displacement) due to an applied force . POW

13 SPRING MOTION

14 Hooke’s Law The amount of force acting on a spring is directly proportional to its change in length as long as it doesn’t exceed its elastic limit.

15 Fs = k x Fs –force on spring (N)
x- change in length (m) k- spring constant (N/m) (spring constant will vary with each spring)

16 The slope of the line is equal to the spring constant
GRAPH The slope of the line is equal to the spring constant Tight spring F(N) Loose spring X (m) Slope=ΔF/Δx=k

17 ELASTIC POTENTIAL ENERGY
k =N/m PEs- joules x = m PE (J) PES = ½ kx2 x (m)

18 F=mg F= .8kg(9.81m/s2) F= 7.85N F=kx 7.85N = k (0.1m) K= 78.5 N/m
Ex: What is the gain in potential energy of a spring that is stretched 10cm by an 800g mass? F=mg F= .8kg(9.81m/s2) F= 7.85N F=kx 7.85N = k (0.1m) K= 78.5 N/m PEs = 1/2kx2 PEs = ½(78.5N/m)(0.1m)2 PEs = Joules

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