Recipe

Compute the energy of a vibrating string

Use when: you're given $u(x,t)$ , or its mode coefficients, for a string satisfying $u_{tt}=c^2u_{xx}$ , and asked for $E(t)$

Goal: the total energy $E(t)=\frac12\int_0^L(u_t^2+c^2u_x^2)\,dx$ at the time you need it

Taught in Energy of a vibrating string

Quick recipe

Decide whether you already have the mode data or a plain formula for u
Differentiate $u$ once in $t$ and once in $x$
Square, scale by $c^2$ , add, and integrate over $[0,L]$
If more than one mode is present, sum the individual mode energies

1
Check whether the solution is already broken into normal modes with known $A_n,B_n,\omega_n$ , or given only as a single formula for $u(x,t)$ that you'd have to differentiate yourself.
- the mode coefficients are already known — skip differentiating, go straight to the closed-form formula in Step 3.
- only $u(x,t)$ itself is given — differentiate it directly, continuing to Step 2.
Why: the closed-form per-mode energy only exists once the mode decomposition is already in hand; without it, the definition itself is the only route.
2
Compute $u_t(x,t)$ and $u_x(x,t)$ at whatever time you need the energy, and evaluate them there before squaring anything.

Example

For $u(x,t)=[4\cos(2t)+3\sin(2t)]\sin(x)$ on $L=\pi$ : $u_t(x,0)=6\sin(x)$ and $u_x(x,0)=4\cos(x)$ .

Why: the definition needs the velocity and the slope separately, since one feeds the kinetic term and the other the potential term.
3
Multiply $u_x^2$ by $c^2$ , add it to $u_t^2$ , integrate the sum from $0$ to $L$ , and multiply the result by $\tfrac12$ .

Example

$E(0)=\frac12\int_0^\pi\big[(6\sin x)^2+4(4\cos x)^2\big]dx=\frac12\big[36(\tfrac\pi2)+64(\tfrac\pi2)\big]=25\pi$ .

⚡ Shortcut

If the solution is already a single mode $u_n=[A_n\cos\omega_nt+B_n\sin\omega_nt]\sin(n\pi x/L)$ , skip the integral entirely: $E_n=\frac{L\omega_n^2}4(A_n^2+B_n^2)$ . For the same mode ( $A_1=4,B_1=3,\omega_1=2,L=\pi$ ), $E_1=\frac{\pi(4)}4(16+9)=25\pi$ , matching the direct integral exactly.

Why: kinetic and potential energy densities are, by definition, exactly these two squared and scaled terms added together.
4
Compute each mode's own energy from Step 3 separately, then add them. Don't integrate the full multi-mode $u$ as one combined expression.

Example

Adding a second mode $B_2=5,\omega_2=4$ to the string above gives $E_2=\frac{\pi(16)}4(25)=100\pi$ , so the total is $E=25\pi+100\pi=125\pi$ .

Why: cross terms between different modes integrate to zero by the same orthogonality that makes the Fourier sine series work, so nothing is lost by adding the modes' energies separately.

Common mistakes

Using $u(x,t)$ itself in the integral instead of $u_t$ and $u_x$ ; that measures a different quantity entirely, not the energy.
Forgetting the $c^2$ factor on the potential term, or attaching it to the kinetic term instead.
Integrating a multi-mode $u$ as one expression and expecting the cross terms to matter, instead of summing each mode's energy separately.
Dropping the leading $\tfrac12$ in front of the whole integral.