Differential-Equations · Unit 10 · Video 2 · Interactive Practice

When Does a Damped Spring Oscillate? The Condition p < ω₀

IKey Formulas

Formula	Name	Where it comes from
$y'' + 2p\,y' + \omega_0^2\,y = 0$	Standard form	$m x'' + c x' + k x = 0$ divided by $m$ : $2p = \frac{c}{m}$ , $\omega_0^2 = \frac{k}{m}$
$r = -p \pm \sqrt{p^2 - \omega_0^2}$	Characteristic roots	Quadratic formula on $r^2 + 2p\,r + \omega_0^2 = 0$
$y = A\cos(\omega_0 t - \varphi)$	Undamped solution ( $p = 0$ )	$A = \sqrt{c_1^2 + c_2^2}$ , $\tan\varphi = \frac{c_2}{c_1}$ , period $\frac{2\pi}{\omega_0}$
$p < \omega_0 \iff c^2 < 4mk$	Oscillation condition	Complex roots: $p^2 - \omega_0^2 < 0$

Key Insight: The solutions oscillate exactly when the roots are complex, which is exactly when $p < \omega_0$ . At $p = \omega_0$ (critically damped) and for $p > \omega_0$ (overdamped) the roots are real and the mass never swings back and forth.

IIVisualization 1 — No Damping: One Cosine at Frequency $\omega_0$

With $p = 0$ , every solution $c_1\cos\omega_0 t + c_2\sin\omega_0 t$ is a single cosine $A\cos(\omega_0 t - \varphi)$ with period $2\pi/\omega_0$ .

c₁ 2 c₂ 1.5 ω₀ 2

IIIVisualization 2 — Where the Roots Go as $p$ Grows

For $\omega_0 = 2$ , the roots $-p \pm \sqrt{p^2 - 4}$ are complex exactly when $p < 2$ ; at $p = 2$ they merge.

p 1.00

💡 While the roots are complex their product is $\omega_0^2$ and they are conjugates, so $|r|^2 = p^2 + (\omega_0^2 - p^2) = \omega_0^2$ : they slide along the circle $|r| = \omega_0$ until they meet at $r = -\omega_0$ .

IVVisualization 3 — The Test in the Original Constants

With $p = \frac{c}{2m}$ and $\omega_0^2 = \frac{k}{m}$ , the test $p < \omega_0$ becomes $c^2 < 4mk$ , so mass matters too.

m 1 c 2 k 4

VQuiz Questions

Problem 1 · Roots from the Standard Form

Given: $y'' + 6y' + 25y = 0$ — find the characteristic roots using $r = -p \pm \sqrt{p^2 - \omega_0^2}$ .

✅ Correct!

2p = 6

gives

p = 3

\omega_0^2 = 25

gives

\omega_0 = 5

, and

\sqrt{9 - 25} = \sqrt{-16} = 4i

. Since

p = 3 < \omega_0 = 5

, the solutions oscillate.

❌ Not quite. The middle coefficient is

2p

, not

p

2p = 6

means

p = 3

. Using

p = 6

gives

-6 \pm \sqrt{36 - 25}

, which is the wrong quadratic.

❌ Not quite. The radicand is

p^2 - \omega_0^2 = 9 - 25 = -16

, not

25 - 9 = 16

. A negative radicand gives imaginary parts,

\pm 4i

, not the real numbers

\pm 4

❌ Not quite. The real part is

-p = -3

. The formula starts with

-p

because the quadratic formula gives

-\frac{2p}{2}

Show solution

Match $y'' + 6y' + 25y = 0$ to $y'' + 2p\,y' + \omega_0^2 y = 0$ : $2p = 6$ , so $p = 3$ ; $\omega_0^2 = 25$ , so $\omega_0 = 5$ .

r = -p \pm \sqrt{p^2 - \omega_0^2} = -3 \pm \sqrt{9 - 25} = -3 \pm \sqrt{-16} = -3 \pm 4i

Check: $(r + 3)^2 = -16$ expands to $r^2 + 6r + 9 + 16 = r^2 + 6r + 25 = 0$ . ✓

Problem 2 · Does This Spring Oscillate?

Given: $2x'' + 6x' + 5x = 0$ (so $m = 2$ , $c = 6$ , $k = 5$ ) — classify the motion.

✅ Correct!

p = \frac{c}{2m} = \frac{3}{2}

and

\omega_0^2 = \frac{k}{m} = \frac{5}{2}

, so

p^2 = \frac{9}{4} < \frac{10}{4} = \omega_0^2

p < \omega_0

. Equivalently

c^2 = 36 < 4mk = 40

❌ Not quite. This comes from comparing

\frac{c}{m} = 3

with

\omega_0 \approx 1.58

. But

\frac{c}{m}

2p

; the damping constant in the test is

p = \frac{c}{2m} = 1.5

, which is less than

\omega_0 = \sqrt{2.5} \approx 1.58

❌ Not quite. Critical damping needs

c^2 = 4mk

exactly, but

c^2 = 36

and

4mk = 40

. Since

36 < 40

, the system is on the oscillating side.

❌ Not quite. Undamped means

c = 0

, so

p = 0

. Here

c = 6

, so

p = \frac{3}{2} > 0

and the factor

e^{-3t/2}

shrinks every swing.

Show solution

Divide by $m = 2$ : $x'' + 3x' + \frac{5}{2}x = 0$ , so $2p = 3$ and $\omega_0^2 = \frac{5}{2}$ .

p^2 - \omega_0^2 = \frac{9}{4} - \frac{5}{2} = \frac{9}{4} - \frac{10}{4} = -\frac{1}{4} < 0

The radicand is negative, so the roots $r = -\frac{3}{2} \pm \frac{1}{2}i$ are complex and the motion is underdamped. In the original constants: $c^2 = 36 < 4mk = 4 \cdot 2 \cdot 5 = 40$ .

Problem 3 · How Strong a Dashpot Still Lets It Swing?

Given: $x'' + c\,x' + 9x = 0$ (so $m = 1$ , $k = 9$ , $c \ge 0$ ) — find the range of $c$ for which the mass oscillates, and the roots when $c = 4$ .

The mass oscillates exactly when

The characteristic roots when $c = 4$

✅ Correct!

c^2 < 4mk = 36

gives

c < 6

, and

c = 4

is inside that range:

p = 2

\omega_0 = 3

r = -2 \pm \sqrt{4 - 9} = -2 \pm i\sqrt{5}

❌ Not quite. This compares

c

itself with

\omega_0 = 3

. The test uses

p = \frac{c}{2m} = \frac{c}{2}

\frac{c}{2} < 3

gives

c < 6

❌ Not quite. This compares

p = \frac{c}{2}

with

\omega_0^2 = 9

. The test is

p < \omega_0 = 3

, or squared on both sides,

p^2 < \omega_0^2

❌ Not quite. The condition is

c^2 < 4mk = 36

, not

c < 4mk

. Taking the square root gives

c < 6

❌ Not quite. This uses

p = c = 4

. With

m = 1

2p = c = 4

, so

p = 2

; then

p^2 - \omega_0^2 = 4 - 9 = -5

❌ Not quite. The radicand is

p^2 - \omega_0^2 = 4 - 9 = -5

, a difference;

i\sqrt{13}

comes from

-(4 + 9) = -13

, which adds the two squares.

❌ Not quite.

p^2 - \omega_0^2 = 4 - 9 = -5

is negative, so

\sqrt{-5} = i\sqrt{5}

. Real roots would contradict

c = 4 < 6

Show solution

Here $2p = \frac{c}{m} = c$ , so $p = \frac{c}{2}$ , and $\omega_0^2 = \frac{k}{m} = 9$ , so $\omega_0 = 3$ .

Oscillation means $p < \omega_0$ : $\frac{c}{2} < 3$ , so $c < 6$ . The same from $c^2 < 4mk = 4 \cdot 1 \cdot 9 = 36$ .

At $c = 4$ : $p = 2$ and

r = -2 \pm \sqrt{2^2 - 3^2} = -2 \pm \sqrt{-5} = -2 \pm i\sqrt{5}

Check: $(r + 2)^2 = -5$ expands to $r^2 + 4r + 9 = 0$ , the characteristic equation of $x'' + 4x' + 9x = 0$ . ✓

Problem 4 · Make an Overdamped System Swing

Given: $x'' + 4x' + 3x = 0$ ( $m = 1$ , $c = 4$ , $k = 3$ ) is overdamped. Which single change makes the mass oscillate?

✅ Correct! With

m = 2

c^2 = 16 < 4mk = 24

. In standard form

p = \frac{4}{4} = 1

and

\omega_0^2 = \frac{3}{2}

, so

p^2 = 1 < \frac{3}{2} = \omega_0^2

. A heavier mass lowers

p = \frac{c}{2m}

faster than it lowers

\omega_0 = \sqrt{k/m}

❌ Not quite. A stronger dashpot makes things worse:

c^2 = 64

against

4mk = 12

. Oscillation needs

c^2

smaller than

4mk

❌ Not quite. Halving the mass halves

4mk

6

, while

c^2

stays

16

. In standard form

p = \frac{c}{2m}

doubles to

4

but

\omega_0 = \sqrt{k/m}

only grows to

\sqrt{6} \approx 2.45

❌ Not quite. A softer spring lowers

4mk

6

, further below

c^2 = 16

\omega_0

shrinks while

p = 2

stays put.

Show solution

The test is $c^2 < 4mk$ . Originally $c^2 = 16 > 4mk = 12$ (overdamped). Check each change:

$c = 8$ : $64$ vs $12$ — still overdamped.
$m = 2$ : $16$ vs $4 \cdot 2 \cdot 3 = 24$ — $16 < 24$ , oscillates.
$m = \frac{1}{2}$ : $16$ vs $6$ — overdamped.
$k = \frac{3}{2}$ : $16$ vs $6$ — overdamped.

With $m = 2$ the equation is $2x'' + 4x' + 3x = 0$ , i.e. $x'' + 2x' + \frac{3}{2}x = 0$ : $p = 1$ , $\omega_0^2 = \frac{3}{2}$ , roots $r = -1 \pm \sqrt{1 - \frac{3}{2}} = -1 \pm \frac{i}{\sqrt{2}}$ .

Solved: 0 / 4

When Does a Damped Spring Oscillate? The Condition p < ω₀

IKey Formulas

IIVisualization 1 — No Damping: One Cosine at Frequency ω0\omega_0ω0​

IIIVisualization 2 — Where the Roots Go as ppp Grows

IVVisualization 3 — The Test in the Original Constants

VQuiz Questions

IIVisualization 1 — No Damping: One Cosine at Frequency $\omega_0$

IIIVisualization 2 — Where the Roots Go as $p$ Grows