Differential-Equations · Unit 10 · Video 3 · Interactive Practice

Damped Oscillations: How Damping Slows the Pseudo-Frequency

IKey Formulas

Formula	Name	Where it comes from
$r = -p \pm i\,\omega_1$ , $\omega_1 = \sqrt{\omega_0^2 - p^2}$	Underdamped roots	$r^2 + 2pr + \omega_0^2 = 0$ with $p < \omega_0$
$y = A\,e^{-pt}\cos(\omega_1 t - \phi)$	Amplitude-phase form	$e^{-pt}(c_1\cos\omega_1 t + c_2\sin\omega_1 t)$ ; $A$ , $\phi$ from $c_1$ , $c_2$
$t_2 - t_1 = \dfrac{2\pi}{\omega_1}$ , $y\!\left(t + \tfrac{2\pi}{\omega_1}\right) = e^{-2\pi p/\omega_1}\,y(t)$	Pseudo-period	Consecutive downward crossings: the angle $\omega_1 t - \phi$ grows by $2\pi$
$\omega_1^2 + p^2 = \omega_0^2$	Frequency triangle	$p = \dfrac{c}{2m}$ , $\omega_0^2 = \dfrac{k}{m}$

Key Insight: The equation fixes $p$ and $\omega_1$ ; the initial conditions fix $A$ and $\phi$ . With $\omega_0$ fixed, a larger $p$ shortens the leg $\omega_1$ , so the pseudo-period $2\pi/\omega_1$ grows and the crossings spread apart.

IIVisualization 1 — One Pseudo-Period Later

For $y = 3e^{-t/4}\cos(2t + \tfrac{\pi}{4})$ , shifting $t$ by $\pi$ multiplies every value by $e^{-\pi/4}$ .

time t 1.20

IIIVisualization 2 — The Frequency Triangle

With $\omega_0 = \sqrt{5}$ fixed, the legs satisfy $\omega_1^2 + p^2 = 5$ : a longer $p$ forces a shorter $\omega_1$ . The right panel plots only the cosine factor $\cos(\omega_1 t + \tfrac{\pi}{2})$ of the solution $y = A\,e^{-pt}\cos(\omega_1 t + \tfrac{\pi}{2})$ , with the decay left out so the timing is easy to see: its zeros are that solution's crossings.

damping p 2.00

💡 The video's example $y'' + 4y' + 5y = 0$ sits at $p = 2$ : roots $-2 \pm i$ , so $\omega_1 = 1$ and $1^2 + 2^2 = 5$ .

IVVisualization 3 — Who Sets the Four Constants?

For $y'' + c\,y' + 4.0625\,y = 0$ ( $m = 1$ ), the start $y(0)$ , $y'(0)$ moves only $A$ and $\phi$ ; only $c$ moves $p$ and $\omega_1$ .

damping c 0.5 y(0) 2.0 y′(0) −3.0

VQuiz Questions

Problem 1 · Read Off the Pseudo-Period

Given: $y'' + 6y' + 25y = 0$ — find the pseudo-period, the time between consecutive downward crossings of the $t$ -axis.

✅ Correct!

p = 3

\omega_0 = 5

, so

\omega_1 = \sqrt{25 - 9} = 4

and

2\pi/\omega_1 = \pi/2

❌ That is the undamped period.

2\pi/5

uses

\omega_0 = 5

; the damping lowers the frequency to

\omega_1 = \sqrt{\omega_0^2 - p^2}

❌ That is the spacing of neighbouring crossings.

\pi/\omega_1 = \pi/4

goes from a downward crossing to the next upward one; downward to downward takes twice that.

❌ $p$ is the decay rate, not a frequency.

2\pi/3

uses

p = 3

; the crossings are timed by

\omega_1

❌ Not quite. Match

2p = 6

and

\omega_0^2 = 25

, then use

\omega_1 = \sqrt{\omega_0^2 - p^2}

Show solution

Compare with $y'' + 2py' + \omega_0^2 y = 0$ : $2p = 6$ gives $p = 3$ , and $\omega_0^2 = 25$ . Since $p = 3 < 5 = \omega_0$ the roots are complex:

\omega_1 = \sqrt{\omega_0^2 - p^2} = \sqrt{25 - 9} = 4, \qquad r = -3 \pm 4i.

Consecutive downward crossings are one full turn of the angle $\omega_1 t - \phi$ apart:

t_2 - t_1 = \frac{2\pi}{\omega_1} = \frac{2\pi}{4} = \frac{\pi}{2}.

Problem 2 · More Damping, Same Spring

Given: $m\,y'' + c\,y' + k\,y = 0$ with $m = 1$ , $k = 13$ . With $c = 4$ the pseudo-period is $\tfrac{2\pi}{3}$ . The damping is raised to $c = 6$ — find the new pseudo-period.

✅ Correct!

p

grows from

2

3

, so

\omega_1

drops from

3

\sqrt{13 - 9} = 2

and the pseudo-period lengthens from

\tfrac{2\pi}{3}

\pi

❌ The timing does change.

\omega_1 = \sqrt{\omega_0^2 - p^2}

depends on

p = c/(2m)

; only the hypotenuse

\omega_0

stays fixed.

❌ Check $p$ .

p = c/(2m) = 3

, not

6

;

p^2 = 9 < 13 = \omega_0^2

, so the motion is still underdamped.

❌ Sign error.

\omega_1^2 = \omega_0^2 - p^2

, not

\omega_0^2 + p^2 = 22

: damping lowers the pseudo-frequency.

❌ Not quite. Recompute

p = c/(2m)

and

\omega_1 = \sqrt{k/m - p^2}

Show solution

Divide by $m = 1$ : $y'' + c\,y' + 13y = 0$ , so $2p = c$ and $\omega_0^2 = 13$ .

Before: $c = 4$ , $p = 2$ , $\omega_1 = \sqrt{13 - 4} = 3$ , pseudo-period $\tfrac{2\pi}{3}$ (as given).

After: $c = 6$ , $p = 3$ , $\omega_1 = \sqrt{13 - 9} = 2$ , so

\frac{2\pi}{\omega_1} = \frac{2\pi}{2} = \pi.

In the frequency triangle the hypotenuse $\sqrt{13}$ is fixed; the leg $p$ grew from $2$ to $3$ , so the leg $\omega_1$ shrank from $3$ to $2$ and the crossings spread out.

Problem 3 · From Initial Conditions to $A$ , $\phi$ and $t_1$

Given: $y'' + 2y' + 5y = 0$ , $y(0) = 1$ , $y'(0) = 1$ — write the solution as $y = A\,e^{-t}\cos(2t - \phi)$ with $A > 0$ , and find the first downward crossing $t_1 > 0$ .

What is the amplitude A?

What is the phase lag φ?

When is the first downward crossing t₁?

✅ Correct!

y = \sqrt{2}\,e^{-t}\cos(2t - \tfrac{\pi}{4})

, and it first crosses downward when

2t - \tfrac{\pi}{4} = \tfrac{\pi}{2}

, at

t_1 = \tfrac{3\pi}{8}

❌ $A$ is not $y(0)$ .

y(0) = A\cos\phi

, which equals

A

only when

\phi = 0

; here the start is moving up, so

\phi \ne 0

❌ Amplitudes do not add.

A = \sqrt{c_1^2 + c_2^2}

, not

c_1 + c_2

❌ You dropped the $-p\,c_1$ term.

y'(0) = -p\,c_1 + \omega_1 c_2

, so

c_2 = (1 + 1)/2 = 1

, not

\tfrac12

❌ Check your amplitude. Find

c_1 = y(0)

and

c_2

from

y'(0) = -c_1 + 2c_2

, then

A = \sqrt{c_1^2 + c_2^2}

❌ Sign of the lag. With

c_1 = 1

and

c_2 = 1

both positive,

\cos\phi

and

\sin\phi

are positive, so

\phi

is in the first quadrant.

❌ You dropped the $-p\,c_1$ term.

\tan\phi = c_2/c_1

with

c_2 = 1

, not

\tfrac12

❌ Sign of $p\,y(0)$ .

y'(0) = -p\,c_1 + \omega_1 c_2

gives

c_2 = (y'(0) + p\,y(0))/\omega_1 = 1

; subtracting

p\,y(0)

gave

c_2 = 0

❌ Check your phase.

\cos\phi = c_1/A

\sin\phi = c_2/A

❌ Sign of $\phi$ .

\tfrac{\pi}{8}

solves

2t + \tfrac{\pi}{4} = \tfrac{\pi}{2}

; the angle is

2t - \tfrac{\pi}{4}

❌ Divide by $\omega_1$ .

3\pi/4

is the angle

\tfrac{\pi}{2} + \phi

; the time is that angle over

\omega_1 = 2

❌ That is the first upward crossing.

7\pi/8

solves

2t - \tfrac{\pi}{4} = \tfrac{3\pi}{2}

, where the cosine is zero and increasing.

❌ Check your crossing. A downward crossing needs

\omega_1 t - \phi = \tfrac{\pi}{2} + 2\pi n

Show solution

Here $2p = 2$ , $\omega_0^2 = 5$ , so $p = 1$ , $\omega_1 = \sqrt{5 - 1} = 2$ and $y = e^{-t}(c_1\cos 2t + c_2\sin 2t)$ .

y(0) = c_1 = 1, \qquad y'(0) = -c_1 + 2c_2 = 1 \;\Rightarrow\; c_2 = 1.

Amplitude and phase: $A = \sqrt{c_1^2 + c_2^2} = \sqrt{2}$ , and $\cos\phi = \tfrac{1}{\sqrt2}$ , $\sin\phi = \tfrac{1}{\sqrt2}$ , so $\phi = \tfrac{\pi}{4}$ :

y = \sqrt{2}\,e^{-t}\cos\!\left(2t - \tfrac{\pi}{4}\right).

At $t = 0$ the angle is $-\tfrac{\pi}{4}$ . The cosine is next zero and decreasing at angle $\tfrac{\pi}{2}$ :

2t_1 - \tfrac{\pi}{4} = \tfrac{\pi}{2} \;\Rightarrow\; t_1 = \tfrac{3\pi}{8}.

The next downward crossing is $t_1 + \tfrac{2\pi}{2} = \tfrac{11\pi}{8}$ , and the upward crossing between them is at $\tfrac{7\pi}{8}$ .

Problem 4 · Recover the Spring from the Motion

Given: a damped spring with $m = 1$ crosses its rest position going downward every $2$ seconds, and $y(t + 2) = e^{-1}\,y(t)$ for every $t$ — find the spring constant $k$ .

✅ Correct!

\omega_1 = \pi

p = \tfrac12

, and

k = m\,\omega_0^2 = \omega_1^2 + p^2 = \pi^2 + \tfrac14

❌ You treated $\omega_1$ as $\omega_0$ . The crossings give

\omega_1 = \pi

; the spring's

\omega_0

is the hypotenuse, so add

p^2

❌ Wrong side of the triangle.

\omega_0

is the hypotenuse:

\omega_0^2 = \omega_1^2 + p^2

, not

\omega_1^2 - p^2

❌ $2$ s is the downward-to-downward time. Setting

\pi/\omega_1 = 2

halves

\omega_1

; the pseudo-period is

2\pi/\omega_1 = 2

, so

\omega_1 = \pi

❌ Not quite. Get

\omega_1

from the crossings,

p

from the factor

e^{-2\pi p/\omega_1}

, then use the triangle.

Show solution

Pseudo-frequency. Consecutive downward crossings are $2\pi/\omega_1$ apart:

\frac{2\pi}{\omega_1} = 2 \;\Rightarrow\; \omega_1 = \pi.

Decay rate. One pseudo-period multiplies every value by $e^{-2\pi p/\omega_1}$ :

e^{-2\pi p/\pi} = e^{-2p} = e^{-1} \;\Rightarrow\; p = \tfrac{1}{2}.

Spring constant. From the frequency triangle, $\omega_0^2 = \omega_1^2 + p^2$ , and $\omega_0^2 = k/m$ with $m = 1$ :

k = \pi^2 + \tfrac{1}{4} \approx 10.12.

(Also $c = 2mp = 1$ , so the equation is $y'' + y' + (\pi^2 + \tfrac14)\,y = 0$ .)

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