Physics 201: Chapter 14 – Oscillations Dec 8, 2009

1. Physics 201: Chapter 14 – OscillationsDec 8, 2009 • Hooke’s Law • Spring • Simple Harmonic Motion • Energy Physics 201, UW-Madison

2. relaxed position relaxed position relaxed position FX = - kx < 0 FX = -kx > 0 FX = 0 x x x • x  0 • x > 0 x=0 x=0 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 xWhere xis the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”) Physics 201, UW-Madison

3. k m k m k m 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... Physics 201, UW-Madison

4. F = -kx a k m x SHM Dynamics • At any given instant we know thatF = mamust be true. • But in this caseF = -kxand ma = • So: -kx = ma = a differential equation Physics 201, UW-Madison

5. 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! Physics 201, UW-Madison

6. ok SHM Dynamics... • 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! Physics 201, UW-Madison

7. y 1 1 1 2 2 3 3 θ 0 x 4 6 -1 4 6 5 5 SHM Dynamics... The period: T = 2π / ω • But wait a minute...what does angular frequency ωhave to do with moving back & forthin a straight line?? The spring’s motion with amplitude A is identical to the x-component of a particle in uniform circular motion with radius A. Physics 201, UW-Madison

8. k m x Example • 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 Sok = (2 kg) x (20 s -1) 2 = 800 kg/s2 = 800 N/m Physics 201, UW-Madison

9. x(t) = A cos(ωt + φ) v(t) = -ωA sin(ωt + φ) a(t) = -ω2A cos(ωt + φ) k m x 0 Initial Conditions Use “initial conditions” to determine phase 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 sinθ cosθ Physics 201, UW-Madison

10. k m x 0 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 π 2π -A Physics 201, UW-Madison

11. m 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 where Physics 201, UW-Madison

12. z θ L where m Differential equation for simple harmonic motion! d mg The Simple Pendulum... • 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 Physics 201, UW-Madison

13. k m s 0 s L Simple Harmonic Motion: Summary k s 0 m Force: Solution: s = A cos(ωt + φ) Physics 201, UW-Madison

14. The potential energy function Physics 201, UW-Madison

15. t t Energy for simple harmonic motion: E=K+U U K t

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