JEE Mains Physics · Waves

Superposition and Standing Waves on Strings

Two waves at one point add their displacements; two equal waves running in opposite directions on a string fixed at both ends make a standing wave with frequencies nv/2L.

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Why this matters

Fourteen PYQs, thirteen of them asking for a number, and one from 2026. Five find the resultant amplitude of two waves or the phase difference between them, and one counts loud points as a recorder moves between two loudspeakers. Eight are standing waves on a string fixed at both ends: harmonics, a length or a hanging mass for a new frequency, and the amplitude at one point. Almost all are short numerical answers with no options to check against.

Concept 1 of 2: Resultant amplitude of two waves of the same frequency

Two waves of the same frequency add to one wave of that frequency. How big it is depends on the phase difference: in step they add, half a cycle apart they cancel, and in between the amplitudes add like vectors at an angle φ. First write each wave as a single sine with its own phase, then compare the phases.

Definition

$A^{2} = A_1^{2} + A_2^{2} + 2A_1A_2\cos\phi$ .
In phase ( $\phi = 0, 2\pi, \dots$ ): $A = A_1 + A_2$ . Opposite phase ( $\phi = \pi, 3\pi, \dots$ ): $A = |A_1 - A_2|$ .
Equal amplitudes a: $A = 2a\cos\dfrac{\phi}{2}$ .
$a\sin\omega t + b\cos\omega t = \sqrt{a^{2} + b^{2}}\,\sin(\omega t + \alpha)$ with $\tan\alpha = b/a$ : rewrite such a sum as one sine before comparing phases.
A shift $x_0$ inside $\sin k(x - vt + x_0)$ is a phase $kx_0$ ; a constant c inside $\sin 2\pi(x - vt + c)$ is a phase $2\pi c$ .
Path difference $\Delta$ from two sources in step: a maximum when $\Delta = n\lambda$ , a minimum when $\Delta = (n + \tfrac{1}{2})\lambda$ . Moving a detector past N maxima means $\Delta$ changed by $N\lambda$ .

Resultant amplitude

A^{2} = A_1^{2} + A_2^{2} + 2A_1A_2\cos\phi \qquad \phi = \frac{2\pi}{\lambda}\,\Delta

Worked example

Two waves

y_1 = 3\sin\omega t

and

y_2 = 4\cos\omega t

(in cm) meet at a point. Find the resultant amplitude. What would it be if they were in opposite phase instead?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q120Moderate

Example 1 · Waves · Superposition and Standing Waves on Strings

The displacement equations of two interfering waves are given by

y_{1}= 10\sin\left( \omega t +\frac{\pi}{3} \right)cm,y_{2}=

5\lbrack \sin\omega t +\sqrt{3}\cos\omega t\rbrack cm

respectively. The amplitude of the resultant wave is ____cm.

Amplitudes add only in phase

A₁ + A₂ is the largest possible resultant, reached only when the phase difference is zero or a whole number of cycles. At any other phase use the cosine formula.

A constant inside 2π( ) is not the phase

In sin 2π(x − vt + 0.75) the phase is 2π × 0.75 = 3π/2, not 0.75 rad. Multiply the constant by 2π before putting it into the cosine formula.

Drill 5 more on resultant amplitude of two waves of the same frequency

Concept 2 of 2: Standing waves and harmonics of a string fixed at both ends

A wave and its reflection on a string fixed at both ends make a standing wave, with a node at each end. The string can only hold a whole number of half-wavelengths, so only some frequencies fit: the fundamental v/2L and every whole-number multiple of it. Two neighbouring resonances therefore differ by exactly the fundamental.

Definition

Standing wave $y = 2A\cos kx\,\sin\omega t$ : the amplitude at position x is $|2A\cos kx|$ . Nodes are $\lambda/2$ apart; an antinode lies midway.
Fixed at both ends: $L = n\dfrac{\lambda}{2}$ , so $f_n = \dfrac{nv}{2L} = \dfrac{n}{2L}\sqrt{\dfrac{T}{\mu}}$ , n = 1, 2, 3, … All harmonics occur; the nth has n loops.
Consecutive resonances: $f_{n+1} - f_n = \dfrac{v}{2L}$ , the fundamental.
Same string, same tension: $f \propto \dfrac{1}{L}$ . Sonometer with a hanging mass m: $T = mg$ , so $f \propto \sqrt{m}$ .
Fundamental from the wave speed: $\lambda = 2L$ , $v = 2Lf_1$ .

Harmonics of a string

f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}} \qquad f_{n+1} - f_n = \frac{v}{2L} \qquad y = 2A\cos kx\,\sin\omega t

Worked example

A string 50 cm long with

\mu = 2\ \text{g/m}

is fixed at both ends under a tension of 80 N. Find the wave speed and the first three harmonics. Where is the first node away from an end in the third harmonic?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 July 2022 · Q117Moderate

Example 2 · Waves · Superposition and Standing Waves on Strings

A wire of length

30\text{ }cm

, stretched between rigid supports, has it's

n^{\text{th~}}

and

(n + 1)^{\text{th~}}

harmonics at

400\text{ }Hz

and

450\text{ }Hz

, respectively. If tension in the string is

2700\text{ }N

, it's linear mass density is........

kg/m

Frequency goes as the root of the hanging mass

The tension is mg and f ∝ √T, so f ∝ √m. To raise a sonometer's frequency by a factor of 3, the hanging mass must be 9 times larger, not 3 times.

The difference of two resonances is the fundamental

Two neighbouring resonances of a string differ by v/2L. That difference is the first harmonic itself; it does not tell you n until you divide either frequency by it.

A standing wave's amplitude depends on position

In y = 2A cos kx sin ωt, the factor 2A is only the amplitude at an antinode. At any other point the amplitude is |2A cos kx|, and at a node it is zero.

Drill 7 more on standing waves and harmonics of a string fixed at both ends

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Resultant amplitude of two waves of the same frequency
Resultant amplitude
$A^{2} = A_1^{2} + A_2^{2} + 2A_1A_2\cos\phi \qquad \phi = \frac{2\pi}{\lambda}\,\Delta$
Standing waves and harmonics of a string fixed at both ends
Harmonics of a string
$f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}} \qquad f_{n+1} - f_n = \frac{v}{2L} \qquad y = 2A\cos kx\,\sin\omega t$

Watch out for (5)

Amplitudes add only in phase
→ Resultant amplitude of two waves of the same frequency
A constant inside 2π( ) is not the phase
→ Resultant amplitude of two waves of the same frequency
Frequency goes as the root of the hanging mass
→ Standing waves and harmonics of a string fixed at both ends
The difference of two resonances is the fundamental
→ Standing waves and harmonics of a string fixed at both ends
A standing wave's amplitude depends on position
→ Standing waves and harmonics of a string fixed at both ends

Test yourself on Waves

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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