1. Physics 201: Chapter 14 – Oscillations Dec 8, 2009
Hooke’s Law
Spring
Simple Harmonic Motion
Energy
12/7/2018
Physics 201, UW-Madison

2. Springs
Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.
FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”)
relaxed position
FX = 0 x
x=0
relaxed position
FX = -kx > 0 x
x  0
FX = - kx < 0 x
x > 0
relaxed position
x=0
12/7/2018
Physics 201, UW-Madison

3. Simple Harmonic Motion (SHM)
We know that if we stretch a spring with a mass on the end and let it go, the mass will oscillate back and forth (if there is no friction).
This oscillation is called Simple Harmonic Motion, and is actually easy to understand... k m k m k m
12/7/2018
Physics 201, UW-Madison

4. SHM Dynamics F = -kx a k m x a differential equation
At any given instant we know that F = ma must be true.
But in this case F = -kx and ma =
So: -kx = ma =
a differential equation
12/7/2018
Physics 201, UW-Madison

5. The angular frequency: ω
SHM Dynamics...
define
The angular frequency: ω
The peroid: T = 2π / ω
Try the solution x = A cos(ωt)
This works, so it must be a solution!
12/7/2018
Physics 201, UW-Madison

6. SHM Dynamics... ok We just showed that (from F = ma)
has the solution x = A cos(ωt) but … x = A sin(ωt) is also a solution.
The most general solution is a linear combination (sum) of these two solutions! ok
12/7/2018
Physics 201, UW-Madison

7. SHM Dynamics... The period: T = 2π / ω
But wait a minute...what does angular frequency ω have to do with moving back & forth in a straight line ?? x y -1 1 2 3 4 5 6 θ
The spring’s motion with amplitude A is identical to the
x-component of a particle in uniform circular motion with
radius A.
12/7/2018
Physics 201, UW-Madison

8. Example vMAX = ωA Also: k = mω2
A mass m = 2 kg on a spring oscillates with amplitude A = 10 cm. At t = 0 its speed is maximum, and is v = +2 m/s.
What is the angular frequency of oscillation ω?
What is the spring constant k?
vMAX = ωA
Also:
k = mω2
So k = (2 kg) x (20 s -1) 2 = 800 kg/s2 = 800 N/m k x m
12/7/2018
Physics 201, UW-Madison

9. Initial Conditions Use “initial conditions” to determine phase φ!
x(t) = A cos(ωt + φ)
v(t) = -ωA sin(ωt + φ)
a(t) = -ω2A cos(ωt + φ)
Suppose we are told x(0) = 0 , and x is
initially increasing (i.e. v(0) = positive):
x(0) = 0 = A cos(φ) φ = π/2 or -π/2
v(0) > 0 = -ωA sin(φ) φ < 0 So
φ = -π/2 θ π 2π k x m
cosθ
sinθ
12/7/2018
Physics 201, UW-Madison

10. Initial Conditions... So we find φ= -π/2 x(t) = A cos(ωt - π/2 )
v(t) = -ωA sin(ωt - π/2 )
a(t) = -ω2A cos(ωt - π/2 )
x(t) = A sin(ωt)
v(t) = ωA cos(ωt)
a(t) = -ω2A sin(ωt) A
x(t) ωt k x m π 2π -A
12/7/2018
Physics 201, UW-Madison

11. What about Vertical Springs?
We already know that for a vertical spring
if y is measured from the equilibrium position
So this will be just like the horizontal case: j k
y = 0
Which has solution y = A cos(ωt + φ)
F = -ky m
where
12/7/2018
Physics 201, UW-Madison

12. The Simple Pendulum... z θ L where m
Recall that the torque due to gravity about the rotation (z) axis is
τ = -mgd
d = Lsin θ ≈ Lθ for small θ
so τ = -mg Lθ
But τ = Iα, where I = mL2 θ L d m mg z
where
Differential equation for simple harmonic motion!
12/7/2018
Physics 201, UW-Madison

13. Simple Harmonic Motion: Summaryk s m
Force: k s m s L
Solution:
s = A cos(ωt + φ)
12/7/2018
Physics 201, UW-Madison

14. The potential energy function
12/7/2018
Physics 201, UW-Madison

15. Energy for simple harmonic motion: E=K+U
12/7/2018

16. Energy conservation m x x=0 Etotal = 1/2 Mv2 + 1/2 kx2 = constant
KE PE
KEmax = Mv2max /2 = Mω2A2 /2 =kA2 /2
PEmax = kA2 /2
Etotal = kA2 /2
12/7/2018
Physics 201, UW-Madison