Chapter 4 & 5 Oscillator with external forcing-I & II
Problem 5: A mildly damped oscillator driven by an external force is known to have a resonance at an angular frequency somewhere near ω = 1 MHz with a quality factor of 1100. Further, for the force (in Newtons) F (t) = 10 cos(ωt) the amplitude of oscillations is 8.26mm at ω = 1.0 KHz and 1.0 µm at 100MHz. a. What is the spring constant of the oscillator? b. What is the natural frequency 0 of the oscillator? c. What is the FWHM? d. What is the phase difference between the force and the oscillations at ω = ω0 +FWHM/2? Answer: Form equation (5.4) we have the amplitude |x(t)| =
F m
q
(ω02 − ω 2 )2 + 4β 2 ω 2
.
(1)
Here F = 10, ω0 ∼ 1 MHz and β = ω0 /2Q ∼ 106 /2 · 1100 ∼ 103 . a. ω02 >> ω 2 at ω = 1.0 KHz. We can ignore terms containing ω 2 in the denominator of equation 1. Then we have |x(t)| = 8.26 × 10−3 =
10 . mω02
(2)
So k=1210 N/m . b. ω02 << ω 2 at 100MHz. We can ignore 1st term and 3rd term in the denominator of equation 1. Then we have |x(t)| = 10−6 = m = 10−9 kg. 1
10 mω 2
(3)
So ω0 =
q
k/m = 1.1 MHz .
c. FWHM= 2β = 2ω0 /2Q = 103 Hz . d. From equation (5.5) the phase difference between the force and the oscillations at ω = ω0 +FWHM/2= 1100.5 × 103 Hz is φ = tan
system, x¨ + ω02 x = f cos ωt, with initial conditions, x(0) = x(0) ˙ = 0. Show that near resonance, ω → ω0 , x(t) ≈
f t sin ω0 t, 2ω0
that is the amplitude of the oscillations grow linearly
with time. Plot the solution near resonance. (Hint: Take ω = ω0 − ∆ω and expand the solution taking ∆ω → 0.) Answer: Using equation (4.5) we can write the general solution of x¨ + ω02 x = f cos ωt
(5)
x(t) = xc (t) + xp (t) = A cos(ω0 t + φ1 ) + B cos(ωt + φ2 )
(6)
as
Plugging this general solution into equation (5) we get B =
f ω02 −ω 2
and φ2 = 0. Using initial
f conditions, x(0) = x(0) ˙ = 0, we have A = − ω2 −ω 2 and φ1 = 0. Then the final solution is 0
x(t) =
ω02
f (cos ωt − cos ω0 t) − ω2
2
(7)
50 40 30 20
x(t)
10 0 -10 -20 -30 -40 -50 0
5
10
15
20
25
30
35
40
45
t
Figure 1: Solution x(t) (equation (10)) near resonance At near resonance ω = ω0 − ∆ω x(t) =
2f (ω + ω0 )t (ω − ω0 )t sin sin (ω0 + ω)(ω0 − ω) 2 2
(8)
(ω − ω0 )t 2f sin ω0 t × (ω0 + ω)(ω0 − ω) 2
(9)
f t sin ω0 t 2ω0
(10)
x(t) =
x(t) ≈
Problem 12: A horizontal spring with a stiffness constant 9 N/m is fixed on one end to a rigid wall. The other end of the spring is attached with a mass of 1 kg resting on a frictionless horizontal table. At t = 0, when the spring-mass system is in equilibrium and is perpendicular to the wall, a force F (t) = 8 cos 5t N starts acting on the mass in a direction perpendicular to the wall. Plot the displacement of the mass from the equilibrium position between t = 0 and t = 2π neatly. Answer: This problem is similar to the previous. In this problem m = 1kg, ω0 = 3 and
3
1 0.8 0.6 0.4
x(t)
0.2 0 -0.2 -0.4 -0.6 -0.8 -1 0
1
2
3
4
5
6
t
Figure 2: Solution x(t) equation (11) ω = 5. From equation (8) we can write the final solution x(t) = sin 4t sin t
