Wave equation — derivation for string: Question bank

The two figures below are your mental anchors: Figure 1 shows the free-body picture of a tiny curved element (why the pulls don't cancel); Figure 2 contrasts a small-slope wave (glides rigidly) with a large-slope wave (distorts). Refer back to them as you work the items.

True or false — justify

A louder (bigger amplitude) wave travels faster along the string?

False. The speed is

v = T / μ

, which contains no amplitude term; amplitude sets energy carried, not how fast the shape moves — see Energy carried by a wave.

A higher-frequency wave outruns a lower-frequency one on the same ideal string?

False. For the ideal string the dispersion relation gives

v = ω / k =

constant for every frequency, so all frequencies travel together at the same speed.

Doubling the tension doubles the wave speed?

False. Since

v \propto T

, doubling

T

multiplies

v

2 \approx 1.41

; you must quadruple the tension to double the speed.

The wave equation is just Newton's second law applied to a tiny piece of string?

True.

F_{y} = d m a_{y}

with

F_{y} = T y_{xx} d x

and

d m = μ d x

gives

y_{tt} = (T / μ) y_{xx}

directly — it's Newton's Second Law in disguise.

If the string is perfectly straight over a segment, does that segment feel zero net transverse force?

True. Straight means slope is constant, so

\partial^{2} y / \partial x^{2} = 0

; with zero curvature the two end-tensions pull along the same line and cancel — no leftover tug (see Figure 1, straight case).

Does a vertical tension component that is nonzero at one end guarantee a net transverse force?

False. Both ends usually have nonzero vertical components; the net force is their difference, which vanishes if both ends have equal slope (straight string).

Can the horizontal components of tension simply be ignored in the derivation?

False. They are not ignored — they are shown to be equal (

\approx T cos θ \approx T

) at both ends and therefore cancel, which is precisely what keeps

T

uniform along the string.

y = f (x + v t)

also a valid solution, not just

f (x - v t)

True. Both satisfy

y_{tt} = v^{2} y_{xx}

;

f (x - v t)

is a shape sliding right at speed

v

g (x + v t)

one sliding left — the general solution is their sum.

Can the derived speed

v

be negative because a wave can move left?

False.

v = T / μ

is a positive magnitude; leftward motion is captured by the sign inside the argument

x + v t

, not by a negative

v

Is the small-angle assumption a mathematical convenience with no physical content?

False. It is a real physical restriction to small slopes (roughly

θ ≪ 1

rad, i.e.

∣ \partial y / \partial x ∣ ≲ 0.1

); drop it and

sin θ \neq = tan θ

, the equation turns nonlinear, and the pulse distorts as it travels (Figure 2, right).

Spot the error

In each item the first line is the flawed statement and the reveal after ::: says what to inspect and how to fix it. Look for the symbol or sign that Figure 1 would flag.

"Net force

= T sin θ (x + d x) + T sin θ (x)

, adding both vertical pulls."

Error is the $+$ sign. The left end pulls down-and-left, so its vertical component is

- T sin θ (x)

; the correct net force is the difference

T [sin θ (x + d x) - sin θ (x)]

, not the sum.

sin θ \approx tan θ = \frac{\partial y}{\partial t}

is the slope of the string."

Error is the $\partial t$ . Slope is a spatial rate:

tan θ = \partial y / \partial x

, not

\partial y / \partial t

. The time derivative is the transverse velocity, a completely different quantity.

"The net force is

T \frac{\partial y}{\partial x} d x

, proportional to slope."

Error is using the first derivative. The net force is proportional to curvature

T \partial^{2} y / \partial x^{2} d x

; a mere slope means the string is straight-but-tilted and feels no net transverse force — you need the slope to change.

"Since

d m = μ d x

appears on the right and

d x

on the left, the result depends on how small we chose

d x

Error is thinking $d x$ survives. The single power of

d x

cancels from both sides, so the wave equation is independent of the element size — that's exactly why it holds for the whole string.

"Units:

T / μ = N / (kg \cdot m)

, which is not a speed."

Error is the $μ$ units.

μ

is kg per metre (

kg/m

), so

T / μ = (kg\cdotm/s^{2}) / (kg/m) = m^{2} / s^{2}

; the square root is m/s — a genuine speed.

y = A sin (k x - ω t)

solves the wave equation for

any ω and k

Error is 'any'. It works only when

ω = v k

; plugging in gives

- ω^{2} y = - v^{2} k^{2} y

, forcing the dispersion relation

ω = v k

Why questions

Why does curvature — and not slope — create the transverse force?

A constant slope means both end-tensions point along one line and cancel; only a changing slope (curvature) makes the two pulls point differently, leaving a net sideways tug (Figure 1).

Why must we use partial derivatives

\partial y / \partial x

and

\partial y / \partial t

instead of ordinary

d / d x

y (x, t)

depends on two independent variables;

\partial / \partial x

freezes time to read the shape,

\partial / \partial t

freezes position to read the motion — see Partial derivatives and curvature.

Why does the wave speed depend on the medium (

T, μ

) but a sound-related "loudness" does not?

v

comes from the force-per-curvature (

T

) versus inertia-per-length (

μ

) balance — both properties of the string itself, independent of how you shake it.

Why is

T / μ

the natural combination, from a "tug versus sluggishness" view?

T

is the restoring pull (bigger

T

⇒ snappier return ⇒ faster),

μ

is the inertia resisting motion (bigger

μ

⇒ slower); speed grows with

T

and shrinks with

μ

, and the square root makes the units work — Wave speed on a string — v = sqrt(T over mu).

Why can two opposite-moving solutions be added to form standing waves?

The wave equation is linear, so any sum of solutions is a solution; a right-mover plus an equal left-mover interfere into fixed nodes and antinodes — Standing waves on a string.

Why does raising a guitar string's pitch require a lot of tightening for a little gain?

Pitch

\propto v \propto T

, so it grows only with the square root of tension; to raise the frequency by a factor of 2 you must pull the string four times as hard.

Why does the same

F = ma

derivation not care whether the wave is transverse or longitudinal in general?

The method (net force from spatial variation, then

ma

) is universal, but the physical restoring quantity differs — tension here versus compression elsewhere; contrast in Transverse vs Longitudinal Waves.

Edge cases

What happens to the derivation if the displacement is not small (large amplitude)?

sin θ \approx tan θ

fails and

cos θ \neq \approx 1

, so tension is no longer uniform; the equation becomes nonlinear and pulses steepen or distort instead of gliding rigidly (Figure 2, right panel).

μ \to 0

(a massless string) with finite tension, what does

v

predict?

v = T / μ \to \infty

; an inertia-free string would transmit disturbances instantaneously — a signal that the idealisation has broken down, since no real string is massless.

T \to 0

(a slack string), what happens to wave propagation?

v = T / μ \to 0

; with no restoring pull there is nothing to snap a bent piece back, so transverse waves cannot travel — the medium can't support them.

A perfectly flat, horizontal string at rest (

y \equiv 0

) — is it a solution of the wave equation?

Yes, the trivial one: both

y_{tt}

and

y_{xx}

are zero, so

0 = v^{2} \cdot 0

holds; equilibrium is always a valid (if boring) solution.

At the exact crest of a sine wave, the slope is zero — is the transverse force there also zero?

No. Slope zero means the shape's peak, but the force follows curvature

y_{xx}

, which is maximum in magnitude (most sharply bent) at the crest — hence the strongest downward tug.

At an inflection point of the waveform, where curvature

y_{xx} = 0

, what is the net transverse force?

Zero at that instant, since force

\propto y_{xx} = 0

; the string is momentarily straight there, so that element is not being accelerated even though it is moving.