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Harmonic Motion (IV)

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ETot = K U = 1/2mv2 1/2kx2. for a spring. But we know that: x=A ... ETot=1/2kA2 = K U = 1/2mv2 1/2kx2. So, if x=0, all E is in kinetic, and v is at max ... – PowerPoint PPT presentation

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Title: Harmonic Motion (IV)


1
Harmonic Motion (IV)
  • Energy of a simple harmonic oscillator

2
Energy of a SHO
  • Recall ETot K U
    1/2mv2 1/2kx2
  • for a spring. But we know that
    xA cos(?tf) v-A ? sin(?tf)

3
Energy in SHM
Look again at the block spring
Hence ETot ½kA2
We could also write E KU ½ m(vmax )2
4
Concept Quiz
Suppose you double the amplitude of the motion,
what happens to the maximum speed ?
  • Doubles
  • 4 x Larger
  • Doesnt change

5
Concept Quiz
  • Suppose you double the amplitude of the motion,
    what
  • happens to the maximum acceleration?
  • Doubles
  • 4 x Larger
  • Doesnt change

6
Concept Quiz
  • Suppose you double the amplitude of the motion,
    what
  • happens to the the total energy?
  • Doubles
  • 4 x Larger
  • Doesnt change

7
Energy
  • Since we know the total energy of a SHM, we can
    calculate the or displacement velocity at any
    point in time ETot1/2kA2 KU 1/2mv2
    1/2kx2
  • So, if x0, all E is in kinetic, and v is at max
    if xA, all E is in potential, and v is zero

8
Example
  • A 100g block is 5cm from the equilibrium position
    moving at 1.5m/s.
  • a) What is the total energy of the system ?
  • b) What is the amplitude of the oscillations ?

9
Example
A 500g block on a spring is pulled 20cm and
released. The motion has a period of 0.8s.
What is the velocity when the block is 15.4cm
from the equilibrium ?
10
Example
  • A 1.0kg block is attached to a spring with
    k16N/m. While the block is at rest, a student
    hits it with a hammer and almost instantaneously
    gives it a speed of 40cm/s.
  • what is the amplitude of the subsequent
    oscillations ?
  • what is the blocks speed at the point where
    xA/2 ?
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