SHM differential equation — solution - x = A cos(ωt + φ)

WHAT is the equation?

WHY this is the equation, from Newton: For a mass on a spring, Hooke's law gives restoring force $F=-kx$ . Newton says $F=m\ddot{x}$ . So $m\ddot{x} = -kx \;\Rightarrow\; \ddot{x} = -\tfrac{k}{m}x.$ Define $\omega^2 \equiv k/m$ (positive, so $\omega$ is real) and you land on $\ddot{x}=-\omega^2 x$ . Every SHM system (pendulum, LC circuit, floating block) collapses to this same shape — that's why we solve it once.

HOW to solve it from scratch (derivation, no guessing)

Formula Energy-trick derivation of the solution

Step 1 — Multiply by $\dot{x}$ to make total derivatives appear. $\dot{x}\,\ddot{x} = -\omega^2 x\,\dot{x}$ Why this step? $\dot{x}\ddot{x} = \frac{d}{dt}(\tfrac12\dot{x}^2)$ and $x\dot{x}=\frac{d}{dt}(\tfrac12 x^2)$ . Multiplying lets us integrate both sides cleanly (this is energy conservation in disguise).

Step 2 — Integrate over $t$ : $\tfrac12 \dot{x}^2 = -\tfrac12\omega^2 x^2 + C \;\Rightarrow\; \dot{x}^2 = \omega^2(A^2 - x^2),$ writing the constant as $C=\tfrac12\omega^2 A^2$ so that $\dot{x}=0$ at $x=\pm A$ . Why? $A$ is then automatically the maximum displacement (amplitude) — the turning point where speed vanishes.

Step 3 — Separate variables: $\frac{dx}{\sqrt{A^2-x^2}} = \pm\,\omega\,dt.$ Why? This integral is the definition of $\arccos$ / $\arcsin$ — the sinusoid is forced to appear, we never assumed it.

Step 4 — Integrate: $\arccos\!\Big(\tfrac{x}{A}\Big) = \omega t + \phi \;\Rightarrow\; \boxed{x(t) = A\cos(\omega t + \phi)}$ The two constants $A$ and $\phi$ are the two integration constants a 2nd-order ODE must have.

Fixing the constants from initial conditions: $x(0)=A\cos\phi,\qquad v(0)=\dot x(0)=-A\omega\sin\phi.$ Solve: $A=\sqrt{x_0^2 + \frac{v_0^2}{\omega^2}}, \qquad \tan\phi = -\frac{v_0}{\omega x_0}.$ Why the $v_0/\omega$ ? Velocity is amplitude $\times \omega$ , so to compare a velocity with a position you must divide by $\omega$ to get the same units (metres).

The companion quantities (all derived, not memorised)

Formula Velocity, acceleration, period

Differentiate $x=A\cos(\omega t+\phi)$ : $v = \dot x = -A\omega\sin(\omega t+\phi), \qquad a=\ddot x = -A\omega^2\cos(\omega t+\phi)=-\omega^2 x.\;\checkmark$

Max speed $v_{\max}=A\omega$ (at $x=0$ , centre — fastest where pull is zero).
Max acceleration $a_{\max}=A\omega^2$ (at $x=\pm A$ , ends — pull is strongest there).
Period: cosine repeats when its argument grows by $2\pi$ : $\omega T = 2\pi \Rightarrow T=\dfrac{2\pi}{\omega}$ , frequency $f=\dfrac{\omega}{2\pi}$ .
Phase-free speed relation (eliminate time): $v=\pm\omega\sqrt{A^2-x^2}$ — straight from Step 2.

Figure — SHM differential equation — solution - x = A cos(ωt + φ)

Worked examples

Worked example Example 1 — Spring released from rest

Mass $m=0.5$ kg, spring $k=200$ N/m, pulled to $x_0=0.1$ m and released from rest ( $v_0=0$ ).

Find $\omega$ : $\omega=\sqrt{k/m}=\sqrt{200/0.5}=20$ rad/s. Why? $\omega^2=k/m$ from Newton.
Find $A$ : $A=\sqrt{x_0^2+v_0^2/\omega^2}=\sqrt{0.1^2+0}=0.1$ m. Why? Released from rest → starts at a turning point → that is the amplitude.
Find $\phi$ : $\tan\phi=-v_0/(\omega x_0)=0$ , and $x_0>0$ so $\cos\phi>0 \Rightarrow \phi=0$ . Why pick 0 not $\pi$ ? $x_0$ positive forces $\cos\phi$ positive.
Answer: $x(t)=0.1\cos(20t)$ m. Max speed $v_{\max}=A\omega=2$ m/s.

Forecast-then-Verify

Recall Forecast first, then open

Q: Double the amplitude $A$ . What happens to (a) period $T$ , (b) max speed? Forecast, then check: (a) $T=2\pi/\omega$ has no $A$ → period unchanged (SHM is isochronous!). (b) $v_{\max}=A\omega$ → doubles. The intuition trap "bigger swing = slower" is wrong: it travels farther but in the same time, so it must move faster.

Common mistakes (Steel-manned)

Flashcards

What differential equation defines SHM?

\ddot{x}+\omega^2 x=0

, i.e.

\ddot x=-\omega^2 x

(acceleration

\propto -

displacement).

Why must the SHM solution be sinusoidal?

It's the only function whose 2nd derivative equals a negative constant times itself.

General solution of

\ddot x=-\omega^2 x

x=A\cos(\omega t+\phi)

, with constants

A

(amplitude) and

\phi

(phase).

Why exactly two arbitrary constants?

It is a 2nd-order ODE; you need initial position and initial velocity to fix the motion.

Express

A

from

x_0,v_0,\omega

A=\sqrt{x_0^2+v_0^2/\omega^2}

Express

\tan\phi

from

x_0,v_0,\omega

\tan\phi=-v_0/(\omega x_0)

Velocity and acceleration from

x=A\cos(\omega t+\phi)

v=-A\omega\sin(\omega t+\phi)

a=-A\omega^2\cos(\omega t+\phi)=-\omega^2 x

Max speed and max acceleration?

v_{\max}=A\omega

(at centre),

a_{\max}=A\omega^2

(at extremes).

Period and frequency in terms of

\omega

T=2\pi/\omega

f=\omega/2\pi

Does period depend on amplitude in ideal SHM?

No —

T=2\pi/\omega

has no

A

(isochronous).

Speed as function of position (no time)?

v=\pm\omega\sqrt{A^2-x^2}

For a spring, what is

\omega

\omega=\sqrt{k/m}

Recall Feynman: explain to a 12-year-old

Imagine a ball tied to a rubber band at the centre of a table. The further you pull it, the harder the band yanks it back. So it overshoots the middle, the band pulls it back the other way, and it keeps swinging side to side forever. If you filmed its position and drew it, you'd get a smooth wave — exactly the shape of a cosine. The wave's height is how far you first pulled it ( $A$ ), and where the wave starts depends on whether you let go from the side or pushed it from the middle (that's $\phi$ ). The cool magic: how wide you pull it doesn't change how long one full wiggle takes.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, SHM ka pura khel ek hi line mein chhupa hai: $\ddot{x}=-\omega^2 x$ . Iska matlab simple hai — jitna door particle center se jaayega, utna hi zyada force use wapas kheechega, aur force ki direction hamesha center ki taraf hogi (isliye minus sign). Spring mein Hooke ka law $F=-kx$ aur Newton ka $F=m\ddot{x}$ milake yahi equation banti hai, jahan $\omega=\sqrt{k/m}$ .

Ab is equation ka solution kya hai? Hum guess nahi karte — derive karte hain. Dono side ko $\dot{x}$ se multiply karke integrate karo, to milta hai $\dot{x}^2=\omega^2(A^2-x^2)$ (yeh actually energy conservation hai). Phir variables separate karke integrate karo to $\arccos$ apne aap aa jaata hai, aur answer nikalta hai $x=A\cos(\omega t+\phi)$ . Yahan $A$ hai amplitude (maximum kitna door jaata hai) aur $\phi$ hai phase (cycle mein kahan se start kiya). Do constants isliye kyunki second-order equation hai — initial position aur initial velocity, dono chahiye.

Important baat jo exam mein puchte hain: $v_{max}=A\omega$ (center pe sabse fast), $a_{max}=A\omega^2$ (corner pe sabse zyada force), aur $T=2\pi/\omega$ . Note karo — period mein $A$ kahin nahi hai! Matlab chahe chhota swing ho ya bada, ek complete oscillation ka time same rahega (ideal SHM). Yeh isochronous property pendulum clocks ka base hai.

Galti se bachna: $\omega$ aur $f$ alag hain ( $\omega=2\pi f$ ), aur $\phi$ ko hamesha 0 mat maan lena — initial conditions se solve karo, $\cos\phi$ aur $v_0$ ke sign dono check karke quadrant decide karo. Bas itna pakka karlo, SHM ka 80% topic clear ho jaata hai.