1. Simple Harmonic Motion – Dynamics and Energy Contents:
Example
Whiteboards

2. Simple Harmonic Motion - Dynamics
F = ma
F = -kx (Not in data packet)
ma = -kx
a = -kx m
Show SHM with graphs – note that acceleration is opposite x
x = xosin(t)
v = x’ = xocos(t)
a = x’’ = -xo2sin(t)
SOOOO:  =
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3. Simple Harmonic Motion - Energy
Ek (max) = 1/2mvo2
Ep (max) = 1/2kxo2
Where they happen
Derive the energy equations:
Ek = 1/2m2(xo2 – x2)
ET = 1/2m2xo2
Ek: max
Ep: max max
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4. Simple Harmonic Motion - Energy
Ek = 1/2m2(xo2 – x2)
ET = 1/2m2xo2
ET – Total Energy
Ek – Kinetic Energy
 – “Angular” velocity
T – Period of motion
x – Position (at some time)
v – Velocity (at some time)
xo – Max Position (Amplitude)
vo – Max Velocity
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5. a) what is its angular velocity?
Energy Example – An SHO has an amplitude of 0.48 m, a mass of 1.12 kg, and a period of 0.86 seconds.
a) what is its angular velocity?
b) what is its maximum kinetic energy?
c) what is its total energy?
d) what is the kinetic energy when it is 0.23 m from equilibrium? What is its potential energy here?
e) write possible equations for its position and velocity
ω = 2π/(0.86 s) = ≈ 7.3 rad/sec
Ek(max) = ½mω2xo2 = ½(1.12 kg)(7.306 rad/sec)2(0.48 m)2 = ≈ 6.9 J
When it has its maximum kinetic energy, it is passing through the middle (equilibrium) so it has no potential energy, so the total energy is the same as the maximum kinetic energy (so 6.9 J)
Ek = ½mω2(xo2 - x2) = ½(1.12 kg)(7.306 rad/sec)2((0.48 m)2 – (0.23 m)2) = J ≈ 5.3 J
Assuming at t = 0 it was passing through equilibrium headed in the positive direction:
x = (0.48 m)sin((7.306 rad/sec)t)
v = (0.48 m)(7.306 rad/sec)cos((7.306 rad/sec)t) because vo = ωxo
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6. Whiteboards:
Energy
1 | 2 | 3 | 4 | 5
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7. What is its total energy? (save this value in your calculator)
An SHO has a mass of kg, an amplitude of m and an angular velocity of 14.7 rad/sec.
What is its total energy? (save this value in your calculator)
Use ET = 1/2m2xo2 W
0.458 J

8. An SHO has a mass of 0. 259 kg, an amplitude of 0
An SHO has a mass of kg, an amplitude of m and an angular velocity of 14.7 rad/sec.
What is its kinetic energy when it is m from equilibrium? What is its potential energy?
Use Ek = 1/2m2(xo2 – x2) W
0.20 J, 0.26 J

9. a) What is its maximum velocity? b) What is its amplitude of motion?
An SHO has a total energy of 2.18 J, a mass of kg, and a period of s.
a) What is its maximum velocity?
b) What is its amplitude of motion?
Use Ek = 1/2mv2
Then  = 2/T
Use Ek (max) = 1/2m2xo2 W
5.88 m/s J, m

10. An SHO a maximum velocity of 3. 47 m/s, and a mass of 0
An SHO a maximum velocity of 3.47 m/s, and a mass of kg, and an amplitude of m. What is its potential energy when it is m from equilibrium?
 = 2/T
Use Ek = 1/2mv2
Use Ek (max) = 1/2m2xo2
Then Use Ek = 1/2m2(xo2 – x2)
Subtract kinetic from max W
0.170 J

11. a) what is its total energy?
A 1250 kg car bounces up and down with the following equation of motion: (in m)
x = 0.170sin(4.42t)
a) what is its total energy?
b) what is its kinetic energy at t = 3.50 s?
Use ET = 1/2m2xo2
Then find x from the equation: (.04007…)
Then use Use Ek = 1/2m2(xo2 – x2) W
353 J, 333 J