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Waves and Oscillations
Stationary Waves
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Superposition principle, formation of stationary waves, nodes and antinodes, characteristics of standing waves, standing waves in strings (fundamental frequency and harmonics), and open and closed organ pipes.
PGT extension
Analytical treatment of reflection from fixed and free boundaries, phase change on reflection, energy density and energy transmission in stationary waves, organ pipes with end corrections, resonance column apparatus, and Melde's experiment.
Prerequisites
Superposition principle of waves
Wave equation and phase relations
Boundary conditions for strings and air columns
Formation and Characteristics of Stationary Waves
Formed when an incident wave interferes with its reflected counterpart of equal amplitude and frequency.
Energy is not propagated along the medium; it is localized between nodes, leading to standing patterns.
All particles between two successive nodes vibrate in phase with each other, though with different amplitudes.
Nodes and Antinodes
Nodes are points of permanent zero displacement and maximum pressure/stress variation.
Antinodes are points of maximum displacement amplitude and minimum pressure/stress variation.
The separation between consecutive nodes or consecutive antinodes is lambda / 2, while between a node and its adjacent antinode it is lambda / 4.
Stationary Waves in Stretched Strings
A string fixed at both ends supports transverse standing waves with nodes at the fixed boundaries.
The fundamental frequency depends on the length of the string, tension, and mass per unit length (linear mass density).
Integer multiples of the fundamental frequency (harmonics) can be excited as normal modes of vibration.
Stationary Waves in Air Columns (Organ Pipes)
Open organ pipes have displacement antinodes at both open ends, producing all harmonics (even and odd).
Closed organ pipes have a displacement node at the closed end and an antinode at the open end, producing only odd harmonics.
End correction accounts for the antinode forming slightly beyond the actual open boundary of the pipe.
Derivations
Analytical Treatment of Stationary Waves
Conditions: Two sinusoidal progressive waves of equal amplitude, frequency, and wavelength traveling in opposite directions.
Let the incident wave be $y_1 = A \sin(\omega t - kx)$ and the reflected wave traveling in the opposite direction be $y_2 = A \sin(\omega t + kx)$.
By the principle of superposition, the resultant displacement is $y = y_1 + y_2 = A [\sin(\omega t - kx) + \sin(\omega t + kx)]$.
Using the trigonometric identity sin(A - B) + sin(A + B) = 2 sin A cos B, obtain y = 2A cos(kx) sin(omega t).
Result: y = [2A cos(kx)] sin(omega t), representing a stationary wave where amplitude depends on position x as $A_s = 2A \cos(kx)$.
Frequencies of Vibrating String Fixed at Both Ends
Conditions: String of length L under tension T with linear mass density mu.
Apply boundary conditions for a string of length L fixed at x = 0 and x = L, which implies nodes at both ends.
The condition for nodes at both ends requires $L = n(\lambda / 2)$, where n = 1, 2, 3, ...
Substitute lambd$a = v / f$ and velocity of transverse wave $v = \sqrt(T / \mu) into the wavelength relation$ to solve for frequency f.
Result: $f_n = (n / 2L) \sqrt(T / \mu)$, for n = 1, 2, 3, ...
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Stationary Wave Equation
$$y = 2A \cos(kx) \sin(\omega t)$$
y = displacement, A = component amplitude, k = wave number, \omega = angular frequency y in m, A in m, k in rad/m, \omega in rad/s
Superposition of two identical waves traveling in opposite directions
Distance measured from a fixed reflecting boundary
Transverse Wave Velocity in String
$$v = \sqrt{\frac{T}{\mu}}$$
v = velocity, T = tension, \mu = mass per unit length v in m/s, T in N, \mu in kg/m
Uniform flexible string
Frequencies in String Fixed at Both Ends
$$f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$$
$f_n = frequency of n-th harmonic$, L = length of string f in Hz, L in m
n = 1, 2, 3, ...
Frequencies in Open Organ Pipe
$$f_n = \frac{n v}{2L}$$
$f_n = frequency$, v = speed of sound in air, L = length of pipe f in Hz, v in m/s, L in m
n = 1, 2, 3, ... (All harmonics present)
Frequencies in Closed Organ Pipe
$$f_n = \frac{n v}{4L}$$
$f_n = frequency$, v = speed of sound in air, L = length of pipe f in Hz, v in m/s, L in m
n = 1, 3, 5, ... (Only odd harmonics present)
End Correction for Organ Pipe
$$e = 0.3 d$$
e = end correction, d = inner diameter of the pipe m
For a single open end
Graphs and experiments
Displacement Profile of Stationary Waves
Draw: A sinusoidal envelope showing standing wave loops with nodes (zero amplitude) and antinodes (maximum amplitude) labeled along a spatial x-axis.
Interpret: Shows the spatial variation of vibration amplitude, demonstrating that nodes and antinodes maintain fixed spatial positions over time.
Exam use: Tested in identifying node-antinode counts and calculating wavelength from given length dimensions.
Melde's Experiment Setup
Draw: A stretched string connected to an electrically driven tuning fork at one end and passing over a frictionless pulley to a mass hanger at the other end.
Interpret: Demonstrates transverse and longitudinal stationary wave modes in strings and relates frequency, tension, and linear mass density.
Exam use: Numerical problems based on tuning fork frequency and resonance loops in parallel or perpendicular configurations.
Resonance Column Apparatus
Draw: A cylindrical glass tube mounted vertically in a water reservoir, with a tuning fork held above the open mouth.
Interpret: Used to determine the velocity of sound in air by finding resonant lengths corresponding to standing wave nodes and antinodes.
Exam use: Direct questions on first and second resonance lengths and end correction calculations.
Solved numericals
TGT core
A steel wire of length 1.0 m and mass 10 g is stretched with a tension of 400 N. Calculate the fundamental frequency of transverse vibrations of the wire.
Find the linear mass density m$u = mass / length = 10 * 10^-3 kg / 1.0 m = 0.01 kg/m$.
A closed organ pipe and an open organ pipe have the same length. If the fundamental frequency of the open pipe is 300 Hz, calculate the fundamental frequency of the closed pipe, accounting for end correction by treating the effective length as L + 0.3d where diameter d = 0.02 L.
For the open pipe, fundamental frequency $f_o = v / (2L_0) = 300 Hz$.
For the closed pipe without end correction, fundamental frequency $f_c = v / (4L_c)$. Since lengths are equal ($L_0 = L_c = L)$, $f_c = f_o / 2 = 150 Hz$.
With end correction, effective length for open pipe is L_eff,$o = L + 2(0.3d) = L + 0.6d$, and for closed pipe is L_eff,$c = L + 0.3d$.
Since frequency is inversely proportional to effective length, f_c / $f_o = L_eff$,o / L_eff,c. Substitute d = 0.02 L to find the exact ratio and compute f_c.
Answer: 148.5 Hz
Assumptions and limitations
Amplitude of component waves is assumed to be identical; in real reflections, partial absorption causes unequal amplitudes, resulting in non-zero minimum displacement at nodes.
The string is assumed perfectly flexible with uniform cross-section and density, ignoring stiffness and bending rigidity.
Common misconceptions
Common error: Energy travels along a stationary wave from one node to the next.
Correct: Energy is stationary and remains confined between adjacent nodes.
Nodes are points of zero displacement where no energy can flow across.
Common error: All particles in a stationary wave vibrate with the same amplitude.
Correct: Amplitude varies continuously from zero at nodes to maximum at antinodes.
The spatial amplitude factor 2A cos(kx) depends on position x.
Common error: Closed organ pipes contain all harmonics like open pipes.
Correct: Closed organ pipes contain only odd harmonics.
The boundary conditions (node at closed end, antinode at open end) exclude even integer multiples.
Quick revision
Stationary waves are formed by the superposition of two identical waves traveling in opposite directions with no net energy transport.
Stationary Wave Equation: y = 2A \cos(kx) \sin(\omega t)
Node Position Condition: x = \frac{n\lambda}{2} \quad (n = 0, 1, 2, \dots)
Must remember
Stationary waves are formed by the superposition of two identical waves traveling in opposite directions.
Energy does not propagate along a stationary wave; it is localized between adjacent nodes.
The distance between consecutive nodes or consecutive antinodes is $\frac{\lambda}{2}$, and between a node and adjacent antinode it is $\frac{\lambda}{4}$.
A string fixed at both ends vibrates with frequencies given by $f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$ where $n = 1, 2, 3, \dots$
Open organ pipes produce all harmonics, whereas closed organ pipes produce only odd harmonics.
Glossary
NodePoints of permanent zero displacement and maximum pressure/stress variation in a stationary wave.
AntinodePoints of maximum displacement amplitude and minimum pressure/stress variation in a stationary wave.
Linear Mass Density ($\mu$)Mass per unit length of a stretched string, measured in kg/m.
End Correction ($e$)Correction applied to the length of an organ pipe because the displacement antinode forms slightly outside the open end ($e \approx 0.3d$).
Open vs Closed Organ Pipes
Parameter
Open Organ Pipe
Closed Organ Pipe
Boundary Conditions
Antinode at both ends
Node at closed end, antinode at open end
Allowed Frequencies ($f_n$)
$f_n = \frac{nv}{2L}$ ($n = 1, 2, 3, \dots$)
$f_n = \frac{nv}{4L}$ ($n = 1, 3, 5, \dots$)
Harmonics Present
All harmonics (even and odd)
Only odd harmonics
Practice Quiz
Choose an option to reveal the explanation.
1. Which of the following conditions is necessary for the formation of stationary waves?
Explanation: Stationary waves are formed when two identical waves (equal amplitude and frequency) travel in opposite directions and interfere.
2. What is the distance between a node and its adjacent antinode in a stationary wave of wavelength $\lambda$?
Explanation: The distance between a node and its adjacent antinode is always $\frac{\lambda}{4}$.
3. In a stationary wave represented by $y = 2A \cos(kx) \sin(\omega t)$, the nodes are located at positions where:
Explanation: Nodes are points of zero displacement, which occurs when the spatial amplitude term $2A \cos(kx) = 0$, meaning $\cos(kx) = 0$.
4. The fundamental frequency of a stretched string fixed at both ends is proportional to:
Explanation: The fundamental frequency is given by $f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, which is directly proportional to $\sqrt{T}$.
5. An open organ pipe of length $L$ resonates in its fundamental mode. What is the wavelength of the wave?
Explanation: In an open organ pipe, the fundamental mode has an antinode at each end and a node in the middle, so $L = \frac{\lambda}{2}$, giving $\lambda = 2L$.
6. Which harmonics are present in a closed organ pipe?
Explanation: Due to boundary conditions (node at closed end, antinode at open end), a closed organ pipe produces only odd harmonics ($1f, 3f, 5f, \dots$).
7. Energy transmission along a stationary wave is:
Explanation: In stationary waves, energy does not propagate along the medium; it remains localized between adjacent nodes.
8. What is the end correction $e$ for a pipe of inner diameter $d$ with a single open end?
Explanation: The empirical end correction for a single open end is given by $e = 0.3d$.
9. All particles between two successive nodes in a stationary wave vibrate:
Explanation: Particles between two successive nodes are on the same side of the loop and vibrate in phase with each other, though their amplitudes vary depending on their position.
10. If the tension in a stretched string is quadrupled, its fundamental frequency will:
Explanation: Since $f \propto \sqrt{T}$, multiplying tension by 4 increases the frequency by a factor of $\sqrt{4} = 2$.
11. What type of wave is produced on a string fixed at both ends?
Explanation: A string fixed at both ends supports transverse standing (stationary) waves.
12. The ratio of the fundamental frequency of an open organ pipe to that of a closed organ pipe of the same length is:
Explanation: For an open pipe, $f_{open} = \frac{v}{2L}$. For a closed pipe, $f_{closed} = \frac{v}{4L}$. The ratio is $\frac{v/2L}{v/4L} = 2:1$.
13. At a displacement node in a stationary wave, the pressure variation is:
Explanation: Nodes of displacement correspond to maximum pressure and stress variation in a stationary wave.
14. When a wave is reflected from a rigid (fixed) boundary, what phase change occurs?
Explanation: Reflection from a rigid/fixed boundary introduces a phase change of $\pi$ radians ($180^\circ$).
15. If the length of a vibrating string is halved while keeping tension and mass per unit length constant, its fundamental frequency will:
Explanation: Since $f \propto \frac{1}{L}$, halving the length doubles the fundamental frequency.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A string of length $L$ and mass per unit length $\mu$ is subjected to tension $T$. The frequency of its third harmonic is given by:
Explanation: The frequency of the $n$-th harmonic in a string fixed at both ends is $f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$. For the third harmonic, $n = 3$, giving $f_3 = \frac{3}{2L}\sqrt{\frac{T}{\mu}}$.
2. A closed organ pipe has a fundamental frequency of $100\text{ Hz}$. What is the frequency of its first overtone?
Explanation: Closed organ pipes produce only odd harmonics ($1f, 3f, 5f, \dots$). The fundamental is $100\text{ Hz}$ ($1f$), and the first overtone is the third harmonic ($3f$), which is $300\text{ Hz}$.
3. The distance between the first node and the third antinode in a stationary wave is:
Explanation: First node is at $x = 0$. Successive antinodes are at $\frac{\lambda}{4}$ (1st), $\frac{3\lambda}{4}$ (2nd), and $\frac{5\lambda}{4}$ (3rd). Thus, the distance is $\frac{5\lambda}{4}$. Alternatively, $0$ to 3rd antinode ($n=2$): $(2(2)+1)\frac{\lambda}{4} = \frac{5\lambda}{4}$.
4. If an open organ pipe has a fundamental frequency $f$, what will be the fundamental frequency if one of its ends is closed?
Explanation: Fundamental frequency of an open pipe is $f = \frac{v}{2L}$. When one end is closed, it becomes a closed pipe with fundamental frequency $f' = \frac{v}{4L} = \frac{f}{2}$.
5. A transverse stationary wave is established in a string by two waves traveling in opposite directions, given by $y_1 = A \sin(\omega t - kx)$ and $y_2 = A \sin(\omega t + kx)$. The maximum particle velocity at an antinode is:
Explanation: The resultant stationary wave equation is $y = 2A \cos(kx) \sin(\omega t)$. At an antinode, $\cos(kx) = 1$, so $y = 2A \sin(\omega t)$. The particle velocity is $\frac{dy}{dt} = 2A\omega \cos(\omega t)$, giving a maximum velocity of $2A\omega$.
6. In Melde's experiment in transverse mode, a string produces 4 loops with a tuning fork of frequency $f$. If the tension is adjusted such that 2 loops are formed with the same tuning fork, the new tension must be:
Explanation: The frequency is given by $f = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$. For a fixed frequency $f$ and length $L$, $n\sqrt{T}$ is constant. Thus, $n_1 \sqrt{T_1} = n_2 \sqrt{T_2} \implies 4\sqrt{T_1} = 2\sqrt{T_2} \implies 2\sqrt{T_1} = \sqrt{T_2} \implies T_2 = 4T_1$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Which of the following properties distinguishes a stationary wave from a progressive wave?
Explanation: Stationary waves do not transfer energy and possess permanent nodes and antinodes, setting them apart from progressive waves.
2. A string of length $1.5\text{ m}$ is fixed at both ends. What is the wavelength of its fundamental mode of vibration?
Explanation: For the fundamental mode of a string fixed at both ends, $L = \frac{\lambda}{2} \implies \lambda = 2L = 2(1.5\text{ m}) = 3.0\text{ m}$.
3. The equation of a stationary wave is given by $y = 0.04 \sin(50\pi t) \cos(\pi x)$, where $x$ and $y$ are in meters and $t$ in seconds. What is the velocity of the component traveling waves?
Explanation: Comparing with $y = 2A \sin(\omega t) \cos(kx)$, we get $\omega = 50\pi$ and $k = \pi$. The wave velocity is $v = \frac{\omega}{k} = \frac{50\pi}{\pi} = 50\text{ m/s}$.
4. What is the frequency of the second overtone of a closed organ pipe of length $L$ and velocity of sound $v$?
Explanation: Closed organ pipes have frequencies $f_n = \frac{nv}{4L}$ for $n = 1, 3, 5, \dots$. The fundamental is $n=1$, the first overtone is $n=3$, and the second overtone corresponds to $n=5$. Thus, $f_3 = \frac{5v}{4L}$.
5. If the mass per unit length of a string is decreased by $75\%$ while keeping the tension constant, the fundamental frequency will:
Explanation: Since $\mu' = \mu - 0.75\mu = 0.25\mu = \frac{\mu}{4}$, the frequency $f \propto \frac{1}{\sqrt{\mu}}$ becomes $\frac{1}{\sqrt{\mu/4}} = 2 \times \frac{1}{\sqrt{\mu}}$, doubling the frequency.
6. In a stationary wave, the ratio of the maximum amplitude of a particle at an antinode to the amplitude of the component progressive waves is:
Explanation: The component waves have amplitude $A$. The stationary wave equation $y = 2A \cos(kx) \sin(\omega t)$ shows that the maximum amplitude at an antinode is $2A$. Thus, the ratio is $2A : A = 2:1$.
7. An open organ pipe resonates with a tuning fork of frequency $330\text{ Hz}$. If the speed of sound is $330\text{ m/s}$, neglecting end correction, the length of the pipe in its fundamental mode is:
Explanation: For an open pipe, $f = \frac{v}{2L} \implies L = \frac{v}{2f} = \frac{330}{2 \times 330} = 0.5\text{ m}$.
8. Which of the following statements about nodes and antinodes is correct?
Explanation: Nodes are points of zero displacement and maximum pressure/stress variation, while antinodes have maximum displacement and minimum pressure variation.
9. The number of loops formed in a string fixed at both ends vibrating in its $n$-th harmonic is:
Explanation: The $n$-th harmonic of a string fixed at both ends contains exactly $n$ loops (antinodes).
10. When a wave reflects from a free boundary (like an open end of a string or longitudinal displacement reflection), what is the phase change?
Explanation: Reflection from a free boundary occurs without any phase change ($0$ phase shift for displacement).