Superposition, Interference and Stationary Waves
Formation and Characteristics
Stationary waves form when two progressive waves of equal amplitude and frequency travel in opposite directions and Superposition occurs.
$$y = [2A \sin(kx)] \cos(\omega t)$$
The displacement is the product of a space-dependent amplitude and a time-dependent oscillation.
$2A \sin(kx)$=Position-dependent amplitude(m)
$k$=Wave number ($2\pi/\lambda$)(rad/m)
$\omega$=Angular frequency(rad/s)
$\sin(kx) = 0$
→Nodes (Zero amplitude)
$\sin(kx) = \pm 1$
→Antinodes (Maximum amplitude)
Nodes: Points of zero displacement that remain permanently at rest. Consecutive nodes are $\lambda/2$ apart.
Antinodes: Points midway between nodes that oscillate with maximum amplitude $2A$. Node-to-antinode distance is $\lambda/4$.
Energy Trapping: Unlike progressive waves, stationary waves do not transfer energy; they localize it between adjacent nodes.
Phase and Energy Localization
Energy alternates between potential and kinetic forms within 'loops', but cannot flow past the Nodes.
Phase Difference: All particles within a single loop are in-phase, while particles in adjacent loops are $180^\circ$ out-of-phase.
Mean Position: All particles (except nodes) cross their equilibrium position simultaneously.
Energy Balance: When amplitude is maximum, energy is purely potential; when passing equilibrium, it is purely kinetic.
Vibrations in Stretched Strings
A string fixed at both ends supports discrete Harmonics where the length $L$ is a multiple of half-wavelengths.
$$f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$$
The frequency depends on tension, length, and the linear mass density of the string.
$f_n$=Frequency of $n^{th}$ harmonic(Hz)
$L$=Length of vibrating string(m)
$T$=Tension in string(N)
$\mu$=Linear mass density (mass/length)(kg/m)
$n = 1$
→Fundamental Frequency (First Harmonic)
Harmonic Series: Frequencies are integer multiples ($f_1, 2f_1, 3f_1...$). The $n^{th}$ harmonic has $n$ loops.
Wavelength Rule: $\lambda_n = 2L/n$. Each higher harmonic halves the previous wavelength.
Tension Relation: $f \propto \sqrt{T}$. Quadrupling tension doubles frequency.
Length Relation: $f \propto 1/L$. Halving the vibrating length doubles the frequency — this is how guitarists raise pitch by pressing frets.
Density Relation: $f \propto 1/\sqrt{\mu}$. Thicker strings vibrate at lower frequencies, which is why bass strings are wound with extra wire.
Dimensional Check: In $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, units: $\frac{1}{m}\sqrt{\frac{N}{kg/m}} = \frac{1}{m}\sqrt{\frac{kg \cdot m/s^2}{kg/m}} = \frac{1}{s}$ = Hz.
Air Columns (Pipes)
Acoustic resonance depends on boundary conditions: Open ends act as antinodes, while closed ends act as nodes.
$$f_{closed} = \frac{(2n-1)v}{4L}$$
Frequency in a closed pipe involves only odd harmonics due to the node-antinode constraint.
$v$=Speed of sound in air(m/s)
$2n-1$=Odd harmonic multiplier(unitless)
$L$=Length of air column(m)
Open Pipe
→$f_n = \frac{nv}{2L}$ (All harmonics present, $n = 1, 2, 3...$)
$L \to \infty$
→Fundamental frequency approaches zero — longer pipes produce deeper notes
Open Pipe: Both ends are antinodes. Supports all harmonics ($n = 1, 2, 3...$). Fundamental wavelength $\lambda_1 = 2L$.
Closed Pipe: Closed end is a node, open end is an antinode. Only odd harmonics ($n = 1, 3, 5...$). Fundamental wavelength $\lambda_1 = 4L$.
Richness Comparison: An open pipe is 'richer in harmonics' because it supports all integer multiples, while a closed pipe skips every even harmonic.
Proportionality: $f \propto 1/L$ for both pipe types. Doubling the pipe length halves the fundamental frequency.
Fundamental Ratio: For the same length, $f_{1,\text{closed}} = \frac{1}{2} f_{1,\text{open}}$ — a closed pipe sounds one octave lower.
Application Strategy
Worked Examples
Problem Statement
A steel wire of length 0.5 m and mass 2.5 g is under a tension of 100 N. Calculate the frequency of the fundamental note.
Linear mass densityWave speed on stringFundamental mode
Strategic Toolkit
$f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$
Step 1
Calculate Linear Mass Density (μ)
Divide total mass by length. Convert grams to kg first.
$\mu = \frac{0.0025\text{ kg}}{0.5\text{ m}} = 0.005\text{ kg/m}$
Step 2
Write the Frequency Formula
For the fundamental mode ($n = 1$) of a string fixed at both ends:
$f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$
Step 3
Substitute Known Values
With $L = 0.5$ m, $T = 100$ N, $\mu = 0.005$ kg/m:
$f_1 = \frac{1}{2(0.5)} \sqrt{\frac{100}{0.005}} = 1 \times \sqrt{20000}$
Step 4
Calculate
Evaluate the expression:
$f_1 = \sqrt{20000} \approx 141.4\text{ Hz}$
Problem Statement
Determine the length of an open organ pipe that resonates at 512 Hz (speed of sound = 344 m/s). Ignore end correction.
Open pipe harmonicsResonance
Strategic Toolkit
$f = \frac{v}{2L}$
Step 1
Rearrange for Length
Isolate $L$ from the fundamental frequency formula for an open pipe.
$L = \frac{v}{2f}$
Step 2
Calculate Numerical Value
Substitute $v = 344$ and $f = 512$.
$L = \frac{344}{2 \times 512} = \frac{344}{1024} \approx 0.336\text{ m}$
Problem Statement
A pipe closed at one end has length 1 m. Find the fundamental frequency and the next two possible harmonics. (Speed of sound = 340 m/s)
Closed pipe boundary conditionsOdd harmonics only
Strategic Toolkit
$f_n = \frac{nv}{4L}, \quad n = 1, 3, 5...$
Step 1
Find Fundamental Frequency
For a closed pipe, $f_1 = v/4L$.
$f_1 = \frac{340}{4 \times 1} = 85\text{ Hz}$
Step 2
Find Next Harmonics (odd only)
Only odd harmonics exist in a closed pipe: $3f_1, 5f_1$.
$f_3 = 3 \times 85 = 255\text{ Hz}, \quad f_5 = 5 \times 85 = 425\text{ Hz}$
Memorize formulas
String Wave Speed
Depends on tension (stiffness) and density (inertia).
Formula
$v = \sqrt{\frac{T}{\mu}}$
Harmonic Series (Open)
Full integer series for strings and open pipes.
Formula
$f_n = \frac{nv}{2L}$
Harmonic Series (Closed)
Odd integers only for pipes closed at one end.
Formula
$f_n = \frac{(2n-1)v}{4L}$
Node-Node Distance
The distance between any two adjacent nodes is always half a wavelength.
Formula
$d_{NN} = \frac{\lambda}{2}$
String Frequency
Fundamental frequency of a string fixed at both ends.
Formula
$f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$
Review
1
Two points on a stationary wave are separated by a distance of exactly half a wavelength (λ/2). If they are not nodes, what is the phase difference between them?
Ans
180 degrees (π radians)
Review
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Use the adjacent-loop phase rule
In a stationary wave, points separated by half a wavelength lie in neighboring loops. Neighboring loops oscillate in opposite directions, so their phase difference is half a cycle.
$$\Delta \phi = 180^\circ = \pi\ \text{rad}$$
2
A string of length L is fixed at both ends and vibrates in its third harmonic. What is the distance between two adjacent nodes?
Ans
L/3
Review
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Relate the third harmonic to wavelength
For a string fixed at both ends, the third harmonic fits three half-wavelengths into the length. That gives $L = 3\lambda/2$, so one half-wavelength is $L/3$.
$$\text{node spacing} = \frac{\lambda}{2} = \frac{L}{3}$$
3
An open organ pipe has a fundamental frequency of 150 Hz. Which of the following frequencies CANNOT be produced by this pipe?
Ans
525 Hz
Review
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Check for an integer harmonic
An open pipe can produce any integer multiple of the fundamental. Only frequencies of the form $n f_1$ are allowed, so test each option against 150 Hz.
$$525 \div 150 = 3.5 \quad \Rightarrow\quad \text{not an integer harmonic}$$
4
The equation of a stationary wave is $y = 10 \sin(\pi x/4) \cos(20\pi t)$. What is the distance between a node and the next adjacent antinode?
Ans
2 units
Review
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Extract the wavelength from the wave number
The spatial factor is $\sin(kx)$ with $k = \pi/4$. Using $\lambda = 2\pi/k$ gives $\lambda = 8$ units. A node to the next antinode is one quarter of a wavelength.
$$\frac{\lambda}{4} = \frac{8}{4} = 2\ \text{units}$$
5
A string fixed at both ends vibrates in its fundamental mode at 200 Hz. If the length is halved and the tension is quadrupled, what is the new fundamental frequency?
Ans
800 Hz
Review
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Apply the string frequency scaling
For a stretched string, the fundamental frequency varies as $f \propto \frac{1}{L}\sqrt{T}$. Halving the length doubles the frequency, and quadrupling the tension adds another factor of 2.
$$f_{new} = 200 \times 2 \times 2 = 800\ \text{Hz}$$
6
An open pipe and a closed pipe have the same fundamental frequency. What is the ratio of their lengths ($L_{open} : L_{closed}$)?
Ans
2 : 1
Review
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Compare the fundamental formulas
For the same fundamental frequency, an open pipe uses $f_1 = v/(2L)$ while a closed pipe uses $f_1 = v/(4L)$. Solving each for length shows the open pipe must be twice as long.
$$\frac{L_{open}}{L_{closed}} = \frac{v/(2f_1)}{v/(4f_1)} = 2$$
7
A wire of mass $m$, length $L$, under tension $T$ vibrates at fundamental frequency $f$. A second identical wire carries tension $4T$. If both wires are joined end-to-end (total length $2L$) and the higher tension $4T$ is applied across the whole system, what is the new fundamental frequency?
Ans
$f$
Review
Tap to reveal
Track the length and tension factors together
Use $f = \frac{1}{2L}\sqrt{T/\mu}$. Doubling the length contributes a factor of $1/2$, while increasing the tension to $4T$ contributes a factor of $\sqrt{4} = 2$. These cancel exactly.
$$f_{new} = \frac{1}{2(2L)}\sqrt{\frac{4T}{\mu}} = f$$
8
An open organ pipe produces two consecutive overtones at 900 Hz and 1200 Hz. What is the fundamental frequency of this pipe?
Ans
300 Hz
Review
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Use the spacing between consecutive harmonics
In an open pipe, adjacent harmonics are separated by one fundamental frequency. Since 900 Hz and 1200 Hz are consecutive overtones, their difference is the fundamental.
$$f_1 = 1200 - 900 = 300\ \text{Hz}$$
9
Two adjacent resonant frequencies of an air column are observed at 400 Hz and 560 Hz. The air column is:
Ans
A closed pipe with fundamental frequency 80 Hz
Review
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Identify the closed-pipe harmonic spacing
Adjacent resonances in a closed pipe are separated by $2f_1$ because only odd harmonics occur. The observed spacing is 160 Hz, so the fundamental is half of that.
$$2f_1 = 160 \Rightarrow f_1 = 80\ \text{Hz}$$
Question 1 of 9
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1
Two points on a stationary wave are separated by a distance of exactly half a wavelength (λ/2). If they are not nodes, what is the phase difference between them?