Standing waves — formation, nodes, antinodes: Question bank

Before you start, recall the one equation everything hangs on:

True or false — justify

Two identical waves going the same direction can make a standing wave.

False — same direction just gives a bigger travelling wave (still

f (x - v t)

, still moves); you need opposite directions so the

ω t

cancels in the shape factor. See Superposition principle.

A standing wave transports net energy along the string.

False — it is built from two Travelling waves carrying equal energy in opposite directions, so the two flows cancel; energy stays local, swapping kinetic↔potential.

Every point of a standing wave oscillates with the same amplitude.

False — amplitude is

2 A sin (k x)

, which depends on position: zero at nodes,

2 A

at antinodes. Position-dependent amplitude is the definition of "standing".

Adjacent nodes are one wavelength apart.

False —

sin (k x)

crosses zero twice per wavelength, so consecutive nodes are

λ /2

apart.

All points between two adjacent nodes reach their peak at the same instant.

True — they all share the factor

cos (ω t)

, so they hit maximum together; only their maximum size

2 A sin (k x)

differs, not their timing.

Points on opposite sides of a node move in phase.

False — across a node

sin (k x)

flips sign, so those two regions move exactly out of phase (one up while the other is down).

At a node nothing physical is happening.

False — displacement is zero but the slope/strain is maximum there, so the restoring tension peaks; a node is mechanically very active.

A standing wave has a definite speed of propagation.

False — its shape doesn't travel, so there is no propagation speed for the pattern; only the underlying counter-propagating waves have speed

v = ω / k

The frequency of the standing wave equals the frequency of each travelling wave that made it.

True — the time factor is

cos (ω t)

with the same

ω

; superposition adds waves, it does not change their frequency.

Spot the error

"Amplitude of the standing wave is

A

, the same as each incoming wave."

Wrong — at an antinode the amplitude is

2 A

(the two waves add fully at peak times); the constant "

A

" only describes the travelling ingredients.

"Nodes and antinodes are

λ /2

apart, so a node sits right next to an antinode at

λ /2

The

λ /2

spacing is node↔node or antinode↔antinode. A node to its nearest antinode is

λ /4

— halfway between two nodes.

"To find nodes I set

cos (ω t) = 0

Wrong factor — nodes are fixed positions, so you set the space part

sin (k x) = 0

. Setting

cos (ω t) = 0

finds the instants when the whole string is momentarily flat, not the nodes.

x = 0

is always a node because

y = 2 A sin (k x) cos (ω t)

gives

sin (0) = 0

Only true for this particular phase choice. Had the shape factor been

cos (k x)

(different boundary/reflection phase),

x = 0

would be an antinode. The end condition, set by Reflection of waves at boundaries, decides.

"The energy at a node is zero because displacement is zero."

Displacement is zero but a node still stores potential energy in the maximum stretch/strain there, and the tension does mechanical work; energy is not absent, it is elastic rather than kinetic.

"Doubling the amplitude of each incoming wave moves the nodes closer together."

No — node positions depend only on

sin (k x) = 0

, i.e. on

k

(hence

λ

), not on

A

. Amplitude changes how hard antinodes swing, not where the still points are.

Why questions

Why do the

x

and

t

parts separate into

f (x) g (t)

The sum-to-product identity cancels

ω t

inside one factor and

k x

inside the other, leaving

2 A sin (k x) \cdot cos (ω t)

— space and time are no longer tangled in one bracket.

Why can a fixed shape not travel?

Travelling means the shape at a later time is the same shape shifted,

f (x - v t)

; but

sin (k x)

has no

t

inside it, so the shape is nailed to fixed

x

-positions and only its overall size breathes.

Why is a node spaced

λ /2

from the next node but only

λ /4

from an antinode?

sin (k x)

goes zero → peak → zero over a half wavelength; the peak (antinode) sits at the midpoint, so node→antinode is half of

λ /2 = λ /4

Why does reflection at a fixed end create the second wave?

A fixed end can't move, so the returning reflected wave must cancel the incoming one there — that inverted, oppositely-travelling wave is exactly the left-mover needed to build the standing pattern.

Why must a fixed end be a node, not an antinode?

A fixed end is clamped to zero displacement at all times, and only a node has

R (x) = 0

; an antinode swings, which the clamp forbids.

Why do only certain wavelengths form clean standing waves on a bounded string?

The pattern must fit the boundary conditions (node/antinode at each end), so an integer number of half-wavelengths must span the length — this is the origin of normal modes and harmonics.

Why does a node carry maximum restoring force despite zero displacement?

Force comes from the curvature/slope of the string, not the displacement; the string bends most steeply through a node, so the transverse tension component (restoring force) peaks there.

Edge cases

What is the amplitude at

x = 0

when the shape factor is

sin (k x)

Zero —

sin (0) = 0

makes

x = 0

a node; this is the natural pattern for a wave reflected off a fixed end at the origin.

What does a standing wave look like at the instants

cos (ω t) = 0

Momentarily flat everywhere —

y = 0

at every

x

— because the whole shape is scaled by zero; all the energy is kinetic at that instant.

If the two counter-propagating waves have unequal amplitudes (

A_{1} \neq = A_{2}

), do true nodes exist?

No perfect nodes — the smaller wave can't fully cancel the larger, so the minima have nonzero amplitude

∣ A_{1} - A_{2} ∣

; you get a partial standing wave (a standing part plus a leftover travelling part).

At a frequency of zero (

ω = 0

), what happens to the "standing wave"?

cos (0) = 1

freezes the shape as a static

2 A sin (k x)

curve — no breathing, no oscillation; it degenerates into a fixed displacement, not a wave.

Is a single travelling wave alone a standing wave for any special choice of amplitude?

No — one wave always keeps

x

and

t

tangled as

sin (k x - ω t)

; only adding its opposite-direction twin cancels the

ω t

in the shape factor. No amplitude fixes this.

At an antinode, is the velocity of the string zero or maximum?

It depends on the instant: at maximum displacement its velocity is zero, and as it passes through equilibrium its velocity is maximum (

2 A ω

). Antinode names the largest swing, not a fixed speed.

Recall One-line summary of every trap

Nodes/antinodes are positions (set by $sin (k x)$ , i.e. by $k$ ); amplitude is position-dependent ( $2 A sin (k x)$ ); net energy transport is zero; the shape can't travel because $t$ isn't inside it; and boundaries decide which wavelengths survive.

Connections

Standing waves — formation, nodes, antinodes — the parent this bank interrogates.
Superposition principle — why the two waves simply add.
Travelling waves — the moving ingredients.
Reflection of waves at boundaries — makes the returning wave; forces end nodes.
Resonance and normal modes — why only some $λ$ fit.
Wave number k and wavelength — $k = 2 π / λ$ behind every spacing.