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Physics Lesson 16.14.4 - Angular Frequency

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Welcome to our Physics lesson on Angular Frequency, this is the fourth lesson of our suite of physics lessons covering the topic of Alternating Current. LC Circuits, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Angular Frequency

We can check whether the equation

Q(t) = Qmax ∙ cos⁡(ω ∙ t + φ)

is a solution for the differential equation

L ∙ d2 Q/dt2 + 1/C ∙ Q = 0

The first derivative of equation

Q(t) = Qmax ∙ cos⁡(ω ∙ t + φ)

is

i(t) = dQ/dt = -ω ∙ Qmax ∙ sin⁡(ω ∙ t + φ)

while the second derivative of the same equation is

d2 Q/dt2 = -ω2 Qmax ∙ cos⁡(ω ∙ t + φ)

Hence, we can write

-L ∙ ω2 ∙ Qmax ∙ cos⁡(ω ∙ t + φ) + 1/C ∙ Qmax ∙ cos⁡(ω ∙ t + φ) = 0

Cancelling the term Qmax ∙ cos⁡(ω ∙ t + φ) from the above equation, we obtain

-L ∙ ω2 + 1/C = 0

Rearranging the last equation to isolate the angular frequency ω, we obtain

1/C = L ∙ ω2
ω2 = 1/L ∙ C
ω = 1/L ∙ C

This is the same equation for angular frequency obtained earlier through another method, so the approach is correct. The equation of Q(t) is a solution for the differential equation

L ∙ d2 Q/dt2 + 1/C ∙ Q = 0

As for the electrical and magnetic oscillations, we can write for the electrical energy stored in a LC circuit at a given time

We (t) = Q2/2C = Qmax2/2C ∙ cos2 (ω ∙ t + φ)

while for the magnetic energy stored in the magnetic field of inductor at a given time, we have

WM (t) = L ∙ i2/2 = L ∙ ω2 ∙ Qmax2/2 ∙ sin2 (ω ∙ t + φ)

Substituting ω = 1/L ∙ C, we obtain for the magnetic energy stored in a LC circuit

WM = L ∙ Qmax2/2L ∙ C ∙ sin2 (ω ∙ t + φ)
= Qmax2/2C ∙ sin2 (ω ∙ t + φ)

Example 4

A LC circuit is operating at standard values of mains electricity (240V, 50Hz). Calculate:

  1. The current flowing in the circuit 1.753 s after the switch is turned ON if the resistance in the circuit is 30Ω
  2. The energy stored in the magnetic field of inductor at this instant if its inductance is 0.8H
  3. The maximum charge stored in the capacitor plates

Take the initial phase of oscillations as zero (φ = 0).

Solution 4

Clues:

ΔVmax = 240V
f = 50 Hz
φ = 0
R = 30Ω
L = 0.8H
i(1.753) = ?
Wm (1.753) = ?
Qmax = ?

  1. The maximum current in the circuit is
    imax = ∆Vmax/R
    = 240 V/30Ω
    = 8A
    Thus, the amount of current flowing in the circuit at t = 1.753 s, is
    i(t) = -imax ∙ sin (ω ∙ t + φ)
    i(t) = -imax ∙ sin (2π ∙ f ∙ t + φ)
    i(1.753) = -8 ∙ sin (2 ∙ π ∙ 50 ∙ 1.753 + 0)
    i(1.753) = -8 ∙ sin (175.3 ∙ π)
    i(1.753) = -8 ∙ sin (175.3 ∙ π)
    i(1.753) = -8 ∙ sin (174 ∙ π + 1.3 ∙ π)
    i(1.753) = -8 ∙ sin (1.3 ∙ π)
    i(1.753) = -8 ∙ (-0.80)
    = 6.4A
  2. The energy stored in the inductor at the given instant (t = 1.753 s), is
    WM = L ∙ i2/2
    = (0.8 H) ∙ (6.4A)2/2
    = 16.384 J
  3. Given that the energy stored in the magnetic field of a LC circuit is
    WM (t) = L ∙ i2/2 = L ∙ ω2 ∙ Qmax2/2 ∙ sin2 (ω ∙ t + φ)
    we obtain for the maximum electric charge Qmax stored in the capacitor plates:
    imax2 = ω2 ∙ Qmax2
    or
    Qmax = imax/ω
    = imax/2 ∙ π ∙ f
    = (8A)/2 ∙ 3.14 ∙ (50 Hz)
    = 0.0255 C

You have reached the end of Physics lesson 16.14.4 Angular Frequency. There are 5 lessons in this physics tutorial covering Alternating Current. LC Circuits, you can access all the lessons from this tutorial below.

More Alternating Current. LC Circuits Lessons and Learning Resources

Magnetism Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
16.14Alternating Current. LC Circuits
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
16.14.1LC Oscillations
16.14.2Electrical to Mechanical Analogy between Two Oscillating Systems
16.14.3LC Oscillations - A Quantitative Approach
16.14.4Angular Frequency
16.14.5Potential Difference in a LC Circuit

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