Differential-Equations · Unit 7 · Video 4 · Interactive Practice

Amplitude and Phase Lag of a Sinusoidal Response

IKey Formulas

Formula	Name	Where it comes from
$\left\|\frac{1}{\alpha}\right\| = \frac{1}{\|\alpha\|}, \quad \arg\frac{1}{\alpha} = -\arg\alpha$	Reciprocal in polar form	$\frac{1}{\alpha}\cdot\alpha = 1$ : moduli multiply, angles add
$\frac{1}{1 + i(\omega/k)} = A\,e^{-i\phi}$	Polar form of the coefficient	Triangle with legs $1$ and $\omega/k$
$A = \frac{1}{\sqrt{1 + (\omega/k)^2}}, \quad \phi = \tan^{-1}\frac{\omega}{k}$	Amplitude and phase lag	Inverted hypotenuse; angle at the origin
$y_1 = A\cos(\omega t - \phi) = A\cos\big(\omega(t - \phi/\omega)\big)$	Response to $\cos\omega t$	$\mathrm{Re}\,A\,e^{i(\omega t - \phi)}$ ; time lag $\phi/\omega$

Key Insight: The modulus of the coefficient becomes the amplitude and its angle $-\phi$ becomes the phase lag: $\tilde{y} = A\,e^{i(\omega t - \phi)}$ , so $y_1 = A\cos(\omega t - \phi)$ with $0 \le \phi < \pi/2$ .

IIVisualization 1 — The Reciprocal of $1 + i(\omega/k)$

Taking the reciprocal of $\alpha = 1 + i(\omega/k)$ inverts its length and reverses the sign of its angle.

ω/k 1.25

IIIVisualization 2 — Two Arrows Rotating Together

Multiplying $A\,e^{-i\phi}$ by $e^{i\omega t}$ adds the angles, so the response's real part trails the input's by $\phi$ .

ωt 1.20 ω/k

IVVisualization 3 — How $k$ and $\omega$ Set the Amplitude and the Lag

Raising $k$ shrinks $\omega/k$ , which raises $A$ toward $1$ and shrinks the phase lag $\phi$ toward $0$ .

Input frequency ω 2 Conductivity k 1

VQuiz Questions

Problem 1 · Amplitude and Phase Lag from the Formulas

Given: $y' + 4y = 4\cos 3t$ , whose sinusoidal response is $y_1 = A\cos(3t - \phi)$ — find $A$ and $\phi$ .

What is the amplitude A?

What is the phase lag φ?

✅ Correct! With

\omega/k = 3/4

the triangle is a

3

–

4

–

5

triangle scaled by

\tfrac14

: hypotenuse

\tfrac54

, so

A = \tfrac45

and

\phi = \tan^{-1}\tfrac34 \approx 0.644

rad.

❌ Ratio upside down. The vertical leg is

\omega/k = 3/4

, not

k/\omega = 4/3

; here

\omega = 3

and

k = 4

❌ That is $|\alpha|$ , not $|1/\alpha|$ . The hypotenuse

\sqrt{1 + (3/4)^2} = 5/4

is the modulus of

1 + i(\omega/k)

; the amplitude is its reciprocal.

❌ Missing square root.

\frac{1}{1 + (3/4)^2} = \frac{16}{25}

A^2

; the modulus needs

\sqrt{1 + (\omega/k)^2}

❌ Not quite. Use

A = 1/\sqrt{1 + (\omega/k)^2}

with

\omega/k = 3/4

❌ Ratio upside down.

\tan\phi

is the vertical leg

\omega/k

over the horizontal leg

1

, so

\tan\phi = 3/4

❌ Sign.

-\tan^{-1}\tfrac34

is the angle of the coefficient

A\,e^{-i\phi}

; the phase lag

\phi

itself is positive.

❌ Not quite. The angle at the origin satisfies

\tan\phi = \text{opposite}/\text{adjacent} = (\omega/k)/1

;

\cos\phi

would be

1/(5/4) = 4/5

, not

3/4

Show solution

Here $k = 4$ and $\omega = 3$ , so $\omega/k = 3/4$ and $\alpha = 1 + \tfrac34 i$ .

|\alpha| = \sqrt{1 + \left(\tfrac34\right)^2} = \sqrt{\tfrac{25}{16}} = \tfrac54, \qquad A = \frac{1}{|\alpha|} = \frac45

\phi = \tan^{-1}\frac{\omega}{k} = \tan^{-1}\frac34 \approx 0.644\ \text{rad}

So $y_1 = \tfrac45\cos\!\left(3t - \tan^{-1}\tfrac34\right)$ .

Problem 2 · Reciprocal in Polar Form

Given: the coefficient $\dfrac{1}{1 + i\sqrt{3}}$ (the case $\omega/k = \sqrt{3}$ ) — write it in polar form.

✅ Correct!

1 + i\sqrt3 = 2\,e^{i\pi/3}

; the reciprocal inverts the length to

\tfrac12

and negates the angle to

-\pi/3

❌ Length not inverted. The angle is negated correctly, but

|1/\alpha| = 1/|\alpha| = 1/2

, not

2

❌ Angle not negated.

\arg\frac{1}{\alpha} = -\arg\alpha

: the reciprocal sits below the real axis, at

-\pi/3

❌ Wrong angle.

\tan\phi = \sqrt3/1

gives

\phi = \pi/3

;

\pi/6

comes from the ratio

1/\sqrt3

, legs swapped.

❌ Not quite. Find

|\alpha|

and

\arg\alpha

for

\alpha = 1 + i\sqrt3

, then invert the length and negate the angle.

Show solution

For $\alpha = 1 + i\sqrt3$ the legs are $1$ and $\sqrt3$ :

|\alpha| = \sqrt{1 + 3} = 2, \qquad \arg\alpha = \tan^{-1}\sqrt3 = \frac{\pi}{3}

The reciprocal inverts the length and reverses the angle:

\frac{1}{1 + i\sqrt3} = \frac12\,e^{-i\pi/3}

Check: $\tfrac12 e^{-i\pi/3} = \tfrac14 - \tfrac{\sqrt3}{4}i$ , and $\dfrac{1}{1 + i\sqrt3} = \dfrac{1 - i\sqrt3}{(1 + i\sqrt3)(1 - i\sqrt3)} = \dfrac{1 - i\sqrt3}{4}$ ✓

Problem 3 · Full Response and Its Time Lag

Given: $y' + 2y = 2\cos 2t$ — find the sinusoidal response $y_1$ , then how much time after each input peak the response peaks.

What is the response y₁?

How long after an input peak does the response peak?

✅ Correct!

\omega/k = 1

gives

A = 1/\sqrt2

and

\phi = \pi/4

, and

\cos(2t - \pi/4) = \cos\big(2(t - \pi/8)\big)

shifts the curve right by

\pi/8

❌ That amplitude is $|\alpha|$ .

|1 + i| = \sqrt2

; the amplitude is its reciprocal,

1/\sqrt2

, which is less than

1

❌ Sign of the phase. The coefficient is

A\,e^{-i\phi}

, so the exponent becomes

i(\omega t - \phi)

: the response lags,

2t - \pi/4

❌ Missing square root.

\frac{1}{1 + 1^2} = \frac12

A^2

;

A = 1/\sqrt{2}

❌ Not quite. Read off

k = 2

\omega = 2

, so

\omega/k = 1

, then use

A = 1/\sqrt{1 + (\omega/k)^2}

and

\phi = \tan^{-1}(\omega/k)

❌ That is the lag in radians.

\phi = \pi/4

measures the lag along the cycle in

\omega t

; in time it is

\phi/\omega = (\pi/4)/2

❌ Not quite. Factor out

\omega

\cos(\omega t - \phi) = \cos\big(\omega(t - \phi/\omega)\big)

, so the time lag is

\phi/\omega

Show solution

Matching $y' + ky = k\cos\omega t$ gives $k = 2$ and $\omega = 2$ , so $\omega/k = 1$ :

A = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt2}, \qquad \phi = \tan^{-1}1 = \frac{\pi}{4}

y_1 = \frac{1}{\sqrt2}\cos\!\left(2t - \frac{\pi}{4}\right) = \frac12\cos 2t + \frac12\sin 2t

Check: $y_1' = -\sin 2t + \cos 2t$ , so $y_1' + 2y_1 = -\sin 2t + \cos 2t + \cos 2t + \sin 2t = 2\cos 2t$ ✓

Time lag: $\cos\!\left(2t - \frac{\pi}{4}\right) = \cos\!\big(2(t - \frac{\pi}{8})\big)$ , so every peak of the response comes $\phi/\omega = \pi/8 \approx 0.39$ time units after the matching input peak.

Problem 4 · Raising the Input Frequency

Given: $y' + ky = k\cos\omega t$ with the conductivity $k$ held fixed — decide what happens as the input frequency $\omega$ increases.

What happens to A and φ?

As ω → ∞, what does the time lag φ/ω approach?

✅ Correct! A fast-oscillating outside temperature barely gets in:

A \to 0

, the lag angle approaches a quarter cycle, and since

\phi < \pi/2

the time lag

\phi/\omega < \pi/(2\omega) \to 0

❌ That is raising $k$ . Raising

\omega

with

k

fixed makes

\omega/k

larger, not smaller.

❌ Check the angle. The vertical leg

\omega/k

of the triangle grows while the horizontal leg stays

1

, so the angle

\phi

widens.

❌ Check the amplitude. A larger

\omega/k

makes

\sqrt{1 + (\omega/k)^2}

larger, so its reciprocal

A

gets smaller.

❌ Not quite. Track

\omega/k

through

A = 1/\sqrt{1 + (\omega/k)^2}

and

\phi = \tan^{-1}(\omega/k)

❌ That is the limit of $\phi$ , an angle. The time lag divides it by

\omega

, and

\omega \to \infty

❌ That is the other end. As

\omega \to 0

\tan^{-1}(\omega/k) \approx \omega/k

, so

\phi/\omega \to 1/k

; for large

\omega

the lag shrinks.

❌ Not quite.

\phi

stays below

\pi/2

, so

\phi/\omega

is squeezed below

\pi/(2\omega)

Show solution

With $k$ fixed, raising $\omega$ raises the ratio $\omega/k$ , the vertical leg of the triangle of $1 + i(\omega/k)$ .

Amplitude: the hypotenuse $\sqrt{1 + (\omega/k)^2}$ grows without bound, so $A = \dfrac{1}{\sqrt{1 + (\omega/k)^2}}$ decreases toward $0$ .

Phase lag: the horizontal leg stays $1$ while the vertical leg grows, so $\phi = \tan^{-1}(\omega/k)$ increases toward $\pi/2$ but never reaches it.

Time lag: $0 < \dfrac{\phi}{\omega} < \dfrac{\pi/2}{\omega} \to 0$ as $\omega \to \infty$ .

The lag angle approaches a quarter cycle while the cycles themselves get very short, so the delay measured in time vanishes.

Solved: 0 / 4

Amplitude and Phase Lag of a Sinusoidal Response

IKey Formulas

IIVisualization 1 — The Reciprocal of 1+i(ω/k)1 + i(\omega/k)1+i(ω/k)

IIIVisualization 2 — Two Arrows Rotating Together

IVVisualization 3 — How kkk and ω\omegaω Set the Amplitude and the Lag

VQuiz Questions

IIVisualization 1 — The Reciprocal of $1 + i(\omega/k)$

IVVisualization 3 — How $k$ and $\omega$ Set the Amplitude and the Lag