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Stationary Waves | What Are Stationary Waves | Standing Waves

Stationary Waves | What Are Stationary Waves :- When two coherent waves with equal frequency and amplitude move in opposite directions through the same medium and overlap, they form a resultant wave that does not travel forward or backward. This type of wave is called a stationary (standing) wave, in which the amplitude changes periodically with position. It is termed “stationary” because there is no overall transfer of energy along the medium.

In real situations, stationary waves are produced when a wave reflects from a boundary and overlaps with the incoming wave. The interference between the incident and reflected waves leads to the formation of a stationary wave.

Stationary waves can be of two types :

Transverse stationary waves :
Stationary waves in which particles vibrate perpendicular to the direction of the wave (e.g., waves on a stretched string).
Longitudinal stationary waves :
Stationary waves in which particles vibrate parallel to the direction of the wave (e.g., sound waves in air columns like organ pipes).

Stationary Waves Produced On Reflection From The Free End (Rarer Medium)

Let the incident wave be :

$\displaystyle {{y}_{1}}=A\sin (\omega t-kx)$ …..(1)

Reflected wave (no phase change) :

$\displaystyle {{y}_{2}}=A\sin (\omega t+kx)$ …..(2)

Resultant displacement (By Principle of Superposition) :

$\displaystyle y={{y}_{1}}+{{y}_{2}}=A\sin (\omega t-kx)+A\sin (\omega t+kx)$

Using the identity :

$\displaystyle \sin A+\sin B=2\sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right)$

Final equation of stationary wave :

$\displaystyle y=2A\cos (kx)\sin (\omega t)=R\sin (\omega t)$ …..(3)

The equation (3) represents a stationary wave. Here we note that all particles of the medium perform simple harmonic motion (S.H.M.) with the same frequency as that of the interfering waves. However, unlike a progressive wave, the amplitude is not constant; it varies with position and is given by

$\displaystyle R=2A\cos (kx)$ …..(4)

Condition for Antinode (A) (maximum displacement) : The points of maximum amplitudes are called antinodes. For maximum amplitude,

$\displaystyle \cos (kx)=\pm 1$

$\displaystyle \Rightarrow kx=n\pi$

$\displaystyle \Rightarrow \frac{2\pi }{\lambda }x=n\pi$

$\displaystyle \Rightarrow x=n\frac{\lambda }{2}$

Where n = 0,1,2,3,…

Antinode are obtained at x values given by

$\displaystyle x=0,\frac{\lambda }{2},\frac{2\lambda }{2},\frac{3\lambda }{2},\frac{4\lambda }{2},...$

The distance between two successive antinodes is λ/2.

Condition for Node (N) (zero displacement) : The points where amplitude is zero are called nodes. For zero amplitude,

$\displaystyle \cos (kx)=0$

$\displaystyle \Rightarrow kx=\frac{(2n+1)\pi }{2}$

$\displaystyle \Rightarrow \frac{2\pi }{\lambda }x=\frac{(2n+1)\pi }{2}$

$\displaystyle \Rightarrow x=\frac{(2n+1)\lambda }{4}$

Where n = 0,1,2,3,…

Node are obtained at x values given by

$\displaystyle x=\frac{\lambda }{4},\frac{3\lambda }{4},\frac{5\lambda }{4},\frac{7\lambda }{4},...$

The distance between two successive nodes is λ/2.

🔬 Curio Capsules

(1). At antinodes air pressure and density both are low.

(2). At nodes air pressure and density both are high.

(3). Antinode is always formed at the free end (i.e., at x = 0).

(4). The distance between a node (N) and adjoining antinode (A) is λ/4.

Example 1.

The standing wave y = 2A sinkx cosωt in an elastic medium is the result of superposition of two travelling waves y₁ and y₂. If y₁ = A sin(ωt – kx), determine the wave y₁.

Solution :

Given,

$\displaystyle y=2Asinkxcos\omega t$

Using :

$\displaystyle 2\sin A\cos B=\sin (A+B)+sin(A-B)$

We get,

$\displaystyle y=A\sin (kx+\omega t)+Asin(kx-\omega t)$

By Principle of Superposition :

$\displaystyle y={{y}_{1}}+{{y}_{2}}$

$\displaystyle \Rightarrow {{y}_{2}}=y-{{y}_{1}}$

$\displaystyle \Rightarrow {{y}_{2}}=\left[ A\sin (kx+\omega t)+sin(kx-\omega t) \right]-\left[ Asin(\omega t-kx) \right]$

$\displaystyle \Rightarrow {{y}_{2}}=\left[ A\sin (kx+\omega t)+sin(kx-\omega t) \right]-\left[ -Asin(kx-\omega t) \right]$

$\displaystyle \therefore {{y}_{2}}=A\sin (kx+\omega t)+2Asin(kx-\omega t)$

$\displaystyle {{y}_{2}}=A\sin (\omega t+kx)-2Asin(\omega t-kx)$

Example 2.

The vibrations of a string of length 60 cm fixed at both ends are represented by the equation

$\displaystyle y=4sin\left( \frac{\pi x}{15} \right)cos\left( 96\pi t \right)$

where x and y are in cm and t in seconds.

(i). What is the maximum displacement of a point at x = 5cm ?

(ii). Where are the nodes located along the string ?

(iii). What is the velocity of the particle at x = 7.5 cm and t = 0.25 sec ?

(iv). Write down the equations of the component waves whose superposition gives the above wave.

Solution :

(i). The equation represents a stationary wave having amplitude

$\displaystyle A=4sin\left( \frac{\pi x}{15} \right)$

At x = 5cm,

$\displaystyle A=4sin\left( \frac{5\pi }{15} \right)=4sin\left( \frac{\pi }{3} \right)=4\times \frac{\sqrt{3}}{2}=2\sqrt{3}cm$

(ii). At nodes, the amplitude of the stationary wave is zero, i.e.,

$\displaystyle 4sin\left( \frac{\pi x}{15} \right)=0$

$\displaystyle \Rightarrow \frac{\pi x}{15}=0,\pi ,2\pi ,3\pi ,...$

$\displaystyle \Rightarrow x=0cm,15cm,30cm,45cm,60cm$

(iii). The particle velocity,

$\displaystyle {{\upsilon }_{p}}=\frac{dy}{dt}=4sin\left( \frac{\pi x}{15} \right)\frac{d}{dt}\left[ cos\left( 96\pi t \right) \right]$

$\displaystyle \Rightarrow {{\upsilon }_{p}}=\left( -96\pi \right)4sin\left( \frac{\pi x}{15} \right)sin\left( 96\pi t \right)$

At x = 7.5 cm and t = 0.25 sec,

$\displaystyle {{\upsilon }_{p}}=\left( -96\pi \right)4sin\left( \frac{\pi \times 7.5}{15} \right)sin\left( 96\pi \times 0.25 \right)$

$\displaystyle {{\upsilon }_{p}}=-384sin\left( \frac{\pi }{2} \right)sin\left( 24\pi \right)=-384\times 1\times 0=0$

(iv). Component waves,

$\displaystyle y=4sin\left( \frac{\pi x}{15} \right)cos\left( 96\pi t \right)=2\left[ 2sin\left( \frac{\pi x}{15} \right)cos\left( 96\pi t \right) \right]$

$\displaystyle y=2\left[ sin\left( \frac{\pi x}{15}+96\pi t \right)+sin\left( \frac{\pi x}{15}-96\pi t \right) \right]$

$\displaystyle y=2sin\left( \frac{\pi x}{15}+96\pi t \right)+2sin\left( \frac{\pi x}{15}-96\pi t \right)={{y}_{1}}+{{y}_{2}}$

Hence,

$\displaystyle {{y}_{1}}=2sin\left( \frac{\pi x}{15}+96\pi t \right)\text{ }and\text{ }{{y}_{2}}=2sin\left( \frac{\pi x}{15}-96\pi t \right)$

Stationary Waves Produced On Reflection From Fixed End (Denser Medium)

Let the incident wave be :

$\displaystyle {{y}_{1}}=A\sin (\omega t-kx)$ …..(5)

Reflected wave (π radian phase change) :

$\displaystyle {{y}_{2}}=A\sin (\omega t+kx+\pi )$

$\displaystyle {{y}_{2}}=-A\sin (\omega t+kx)$ …..(6)

Resultant displacement (By Principle of Superposition) :

$\displaystyle y={{y}_{1}}+{{y}_{2}}=A\left[ \sin (\omega t-kx)-\sin (\omega t+kx) \right]$

$\displaystyle y=-2A\sin \left( kx \right)\cos \left( \omega t \right)=R\cos \left( \omega t \right)$ …..(7)

The equation (7) represents a stationary wave. Here also, all particles perform S.H.M. with the same frequency as that of the interfering waves, and the amplitude varies with position as

$\displaystyle R=-2A\sin \left( kx \right)$ …..(8)

Since amplitude is always taken as magnitude,

$\displaystyle \left| R \right|=2A\left| \sin \left( kx \right) \right|$

Condition for Antinode (A) (maximum displacement) :

The points of maximum amplitudes are called antinodes. For maximum amplitude,

$\displaystyle \sin \left( kx \right)=\pm 1$

$\displaystyle \Rightarrow kx=\left( 2n+1 \right)\frac{\pi }{2}$

$\displaystyle \Rightarrow \frac{2\pi }{\lambda }x=\left( 2n+1 \right)\frac{\pi }{2}$

$\displaystyle \Rightarrow x=\frac{\left( 2n+1 \right)\lambda }{4}$

Where n = 0,1,2,3,…

Antinode are obtained at x values given by

$\displaystyle x=\frac{\lambda }{4},\frac{3\lambda }{4},\frac{5\lambda }{4},\frac{7\lambda }{4},...$

The distance between two successive antinodes is λ/2.

Condition for Node (N) (zero displacement) :

The points where amplitude is zero are called nodes. For zero amplitude,

$\displaystyle \sin \left( kx \right)=0$

$\displaystyle \Rightarrow kx=n\pi$

$\displaystyle \Rightarrow \frac{2\pi }{\lambda }x=n\pi$

$\displaystyle \Rightarrow x=n\frac{\pi }{2}$

Where n = 0,1,2,3,…

Node are obtained at x values given by

$\displaystyle x=0,\frac{\lambda }{2},\frac{2\lambda }{2},\frac{3\lambda }{2},...$

The distance between two successive nodes is λ/2.

🔬 Curio Capsules

(1). Node is always formed at the fixed end (x = 0).

(2). The distance between two successive nodes is λ/2. The distance between a node and the nearest consecutive antinode is λ/4. Nodes and antinodes are formed alternately.

(3). In a longitudinal stationary wave, there is maximum variation of pressure, and hence density, at the nodes. There is no variation of pressure and density at the antinodes. Therefore, Antinodes may be called Pressure Nodes and Nodes may be called Pressure Antinodes.

Example 3.

A standing wave is formed by two harmonic waves, y₁ = A sin(ωt – kx) and y₂ =A sin(ωt + kx) tarvelling on a string of density ρ, area of cross-section S, in opposite directions. Find the total mechanical energy between two adjacent nodes on the string.

Solution :

The distance between two adjacent nodes is λ/2 or π/k (∵ k = 2π/λ).

Volume of string between two nodes,

V = (area of cross-section) × (distance between two nodes)

$\displaystyle \Rightarrow V=S\times \frac{\pi }{k}$

Energy density (energy per unit volume) [(Intensity of a Wave | Power of a Wave) , see equation (2)] of a travelling wave is given by,

$\displaystyle u=\frac{1}{2}\rho {{A}^{2}}{{\omega }^{2}}$

Since a standing wave is produced by the superposition of two identical waves travelling in opposite directions, the total energy contained between two successive nodes is twice the energy carried by one progressive wave over a length of $λ /2$ .

E = 2 × (energy stored in a distance of π/k of a travelling wave)

$\displaystyle \Rightarrow E=2\left( u \right)\left( V \right)=2\left( \frac{1}{2}\rho {{A}^{2}}{{\omega }^{2}} \right)\left( S\times \frac{\pi }{k} \right)$

$\displaystyle \Rightarrow E=\frac{\rho {{A}^{2}}{{\omega }^{2}}S\pi }{k}$

Stationary Waves | What Are Stationary Waves | Standing Waves

Stationary Waves Produced On Reflection From The Free End (Rarer Medium)

Stationary Waves Produced On Reflection From Fixed End (Denser Medium)

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