Stationary Waves (Standing wave) Questions in English

(C) When a wave is reflected from a rigid boundary (or fixed end),the reflected wave undergoes a phase change of $\pi$ radians.
This occurs because the rigid boundary exerts an equal and opposite force on the string,resulting in the inversion of the wave pulse.
Therefore,the phase change is $\pi$.

(D) When a wave pulse traveling along a stretched string reaches a fixed end,the fixed end acts as a rigid boundary or a denser medium.
According to the laws of reflection,when a wave is reflected from a fixed boundary,it undergoes a phase change of $\pi$ radians (or $180^o$).
Since the wave is now traveling in the opposite direction,its velocity vector is also reversed.
Therefore,the reflected pulse has a phase change of $180^o$ and its velocity is reversed.

(B) The sound source and its reflection from the wall create a stationary wave pattern due to interference.
Given frequency $f = 170 \, Hz$ and speed of sound $v = 340 \, m/s$.
The wavelength $\lambda$ is given by $\lambda = \frac{v}{f} = \frac{340}{170} = 2 \, m$.
Positions of minimum intensity (nodes) in a stationary wave are separated by a distance of $\frac{\lambda}{2}$.
Therefore,the distance between two adjacent positions of minimum intensity is $\frac{\lambda}{2} = \frac{2}{2} = 1 \, m$.

(C) In a stationary wave,the points of minimum displacement are called nodes $(N)$,and the points of maximum displacement are called antinodes $(A)$.
The distance between two consecutive nodes is $\frac{\lambda}{2}$.
The distance between two consecutive antinodes is $\frac{\lambda}{2}$.
The distance between a node and its nearest antinode is half the distance between two consecutive nodes,which is $\frac{1}{2} \times \frac{\lambda}{2} = \frac{\lambda}{4}$.

(C) In a stationary wave,nodes are points where the displacement of particles is always zero. Because the displacement gradient (which is related to strain) is highest at these points,the pressure variation and strain are maximum at the nodes. Conversely,at antinodes,the displacement is maximum,but the strain is minimum.

(C) In a stationary wave,all particles in a single loop vibrate in phase with each other.
Particles located in adjacent loops are separated by a node.
When two particles are situated on opposite sides of a node,they vibrate in opposite phases.
Therefore,the phase difference between them is $\pi \text{ radians}$,which is equivalent to $180^o$.

(C) The correct answer is $C$.
In a progressive wave,energy is continuously transported from one point to another in the medium as the wave travels.
In contrast,a stationary (or standing) wave is formed by the superposition of two identical waves traveling in opposite directions,resulting in no net transport of energy across any point in the medium.

(B) Stationary waves (also known as standing waves) are formed by the superposition of two waves of the same frequency (or wavelength) and same amplitude traveling in opposite directions along the same path.
Since $v = f \lambda$, if the frequency and wavelength are the same, the speeds are also the same.
Therefore, the condition is that two waves of equal wavelength, equal amplitude, and equal speed travel along the same path in opposite directions.
Thus, option $B$ is the correct answer.

(B) The standard equation of a stationary wave is given by $y = 2a \sin \frac{2\pi x}{\lambda} \cos \omega t$.
Comparing the given equation $y = 5 \sin \frac{\pi x}{3} \cos 40\pi t$ with the standard equation,we get the propagation constant $k = \frac{2\pi}{\lambda} = \frac{\pi}{3}$.
Solving for $\lambda$,we find $\lambda = 6 \ cm$.
The separation between two adjacent nodes in a stationary wave is given by $\frac{\lambda}{2}$.
Therefore,the separation = $\frac{6}{2} = 3 \ cm$.

(D) The given wave equation is $\overrightarrow{\phi}(x, t) = \overrightarrow{j} \sin \left( \frac{2\pi}{\lambda} vt \right) \cos \left( \frac{2\pi}{\lambda} x \right)$.
$1$. Stationary vs Progressive: $A$ wave is stationary if it has nodes (points where the amplitude is always zero). If we set $x = \frac{(2n+1)\lambda}{4}$ for $n = 0, 1, 2, \dots$,the term $\cos \left( \frac{2\pi}{\lambda} x \right)$ becomes zero. Since these points do not depend on time $t$,the amplitude remains zero at these positions. Thus,the wave is a stationary wave.
$2$. Transverse vs Longitudinal: The displacement vector $\overrightarrow{\phi}$ is in the direction of $\overrightarrow{j}$ (the $y$-direction). The wave function depends on $x$,indicating propagation along the $x$-axis. Since the direction of vibration ($y$-axis) is perpendicular to the direction of propagation ($x$-axis),the wave is transverse.
Therefore,the equation represents a transverse stationary wave. The correct option is $(D)$.

(A) In a stationary wave,the displacement of any particle is given by $y = A \sin(kx) \cos(\omega t)$.
For the particles to be at rest,the velocity $v = \frac{dy}{dt} = -A \omega \sin(kx) \sin(\omega t)$ must be zero.
This condition is satisfied when $\sin(\omega t) = 0$,which occurs at $\omega t = 0, \pi, 2\pi, \dots$
Since $\omega = \frac{2\pi}{T}$,the condition $\sin(\omega t) = 0$ implies $t = 0, \frac{T}{2}, T, \dots$
Thus,all particles in the medium come to rest simultaneously twice in every period of oscillation $T$.
Therefore,option $A$ is correct.

(B) stationary wave is formed by the superposition of two waves of the same frequency and amplitude traveling in opposite directions.
Given the incident wave is $y_1 = a \cos (kx - \omega t)$.
For a point $x = 0$ to be a node,the resultant displacement at $x = 0$ must be zero for all time $t$.
Let the second wave be $y_2 = a \cos (kx + \omega t + \phi)$.
The resultant wave is $y = y_1 + y_2 = a [\cos (kx - \omega t) + \cos (kx + \omega t + \phi)]$.
Using the identity $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$,we get:
$y = 2a \cos (kx + \phi/2) \cos (\omega t + \phi/2)$.
At $x = 0$,$y = 2a \cos (\phi/2) \cos (\omega t + \phi/2)$.
For this to be a node (zero displacement) for all $t$,we must have $\cos (\phi/2) = 0$,which implies $\phi/2 = \pi/2$,or $\phi = \pi$.
Substituting $\phi = \pi$ into the equation for the second wave:
$y_2 = a \cos (kx + \omega t + \pi) = -a \cos (kx + \omega t)$.

(D) In a stationary wave,the total energy oscillates between kinetic and potential energy.
When the kinetic energy of the string is maximum,the potential energy is zero.
Potential energy is associated with the deformation (displacement) of the string.
Since potential energy is zero,the displacement of every particle on the string must be zero at that instant.
Therefore,all particles are passing through their mean positions simultaneously.
This results in the string appearing as a straight line along the equilibrium axis.

(B) The standard equation for a stationary wave is given by $y = A \sin(kx) \cos(\omega t)$.
Comparing the given equation $y = 0.15 \sin(5x) \cos(300t)$ with the standard form,we identify the wave number $k = 5 \text{ rad/m}$.
The relationship between the wave number $k$ and the wavelength $\lambda$ is $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$,we get $5 = \frac{2\pi}{\lambda}$.
Therefore,$\lambda = \frac{2\pi}{5} = \frac{2 \times 3.14159}{5} = \frac{6.283}{5} = 1.2566 \text{ m}$.
Rounding to two decimal places,we get $\lambda \approx 1.26 \text{ m}$.

(D) In a stationary wave,antinodes are the points of maximum amplitude of displacement.
At these points,the variation in pressure is minimum because the air particles move with maximum velocity and the density change is minimal.
Conversely,nodes are the points of zero displacement,where the pressure change is maximum.
Therefore,antinodes correspond to maximum displacement and minimum pressure change.

(D) In a stationary wave,all particles between two consecutive nodes oscillate with the same frequency.
Since they are in the same phase,they all reach their mean position at the same time.
However,the amplitude of vibration for each particle depends on its position $x$ relative to the nodes,given by $A(x) = A_0 \sin(kx)$.
Since the velocity of a particle in simple harmonic motion is given by $v = \omega \sqrt{A^2 - y^2}$,at the mean position $(y = 0)$,the velocity is $v = \omega A$.
Because the amplitude $A$ varies for different particles between the nodes,the maximum velocity at the mean position is different for each particle.
Therefore,all particles pass through the mean position at the same time but with different velocities.

(D) Standing waves are formed by the superposition of two waves of the same frequency and speed traveling in opposite directions.
$(a)$ $A$ string clamped at both ends allows for the reflection of waves at the boundaries,creating standing waves.
$(b)$ $A$ string clamped at one end and free at the other also allows for reflection,resulting in standing waves.
$(c)$ When an incident wave hits a wall,it reflects back,and the superposition of the incident and reflected waves creates standing waves.
Therefore,all the given conditions can produce standing waves. The correct option is $D$.

(A) The distance between two consecutive nodes is $\frac{\lambda}{2}$.
In a standing wave with $3$ nodes and $2$ antinodes,there are $2$ loops.
The total length $L$ between the two extreme nodes is given by $L = 2 \times \frac{\lambda}{2} = \lambda$.
Given that the distance between the two atoms (which act as the extreme nodes) is $1.21 \; \mathring{A}$,we have $L = 1.21 \; \mathring{A}$.
Therefore,the wavelength $\lambda = 1.21 \; \mathring{A}$.

(D) In a stationary wave,the distance between a node and its nearest antinode is given by $\frac{\lambda}{4}$.
Given $\frac{\lambda}{4} = 20 \ cm$,therefore the wavelength $\lambda = 80 \ cm$.
The phase difference $\Delta \phi$ between two particles separated by a distance $\Delta x$ is given by the formula $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting the values $\Delta x = 60 \ cm$ and $\lambda = 80 \ cm$:
$\Delta \phi = \frac{2\pi}{80} \times 60 = \frac{120\pi}{80} = \frac{3\pi}{2}$ radians.

(A) The velocity of the wave is given by $v = n\lambda$,where $n$ is the frequency and $\lambda$ is the wavelength.
Given $v = 1200\, m/s$ and $n = 300\, Hz$.
Calculating the wavelength: $\lambda = \frac{v}{n} = \frac{1200}{300} = 4\, m$.
The distance between a node and the adjacent antinode in a stationary wave is always $\frac{\lambda}{4}$.
Therefore,the required distance $= \frac{4\, m}{4} = 1\, m$.

(A) For two waves to produce a stationary wave,they must have the same frequency,same amplitude,and travel in opposite directions along the same line.
Wave $A$ $(z_1 = a\cos(kx - \omega t))$ travels in the positive $x$-direction.
Wave $B$ $(z_2 = a\cos(kx + \omega t))$ travels in the negative $x$-direction.
Both waves $A$ and $B$ have the same amplitude $a$,the same angular frequency $\omega$,and the same wave number $k$,and they propagate along the same axis ($x$-axis).
Wave $C$ $(z_3 = a\cos(ky - \omega t))$ propagates along the $y$-axis,so it cannot form a stationary wave with $A$ or $B$.
Therefore,the superposition of $A$ and $B$ results in a stationary wave: $z = z_1 + z_2 = a[\cos(kx - \omega t) + \cos(kx + \omega t)] = 2a\cos(kx)\cos(\omega t)$.

(A) The given equation for the standing wave is $Y = A \sin(100t) \cos(0.01x)$.
Comparing this with the standard equation of a standing wave $Y = A \sin(\omega t) \cos(kx)$,we identify the angular frequency $\omega = 100 \ rad/s$ and the wave number $k = 0.01 \ rad/m$.
The velocity of the wave $v$ is given by the ratio of angular frequency to wave number:
$v = \frac{\omega}{k} = \frac{100}{0.01} = 10000 \ m/s$.

(B) At a fixed end,a node is always formed.
Given that a node is formed at a distance of $10 \, cm$ from the fixed end,this distance represents the distance between two consecutive nodes.
The distance between two consecutive nodes is given by $\frac{\lambda}{2} = 10 \, cm$.
Therefore,the wavelength $\lambda = 20 \, cm = 0.2 \, m$.
The speed of the wave $v$ is given by the formula $v = f \lambda$,where $f = 100 \, Hz$.
$v = 100 \times 0.2 = 20 \, m/s$.

(C) stationary wave is formed by the superposition of two waves of the same frequency and amplitude traveling in opposite directions.
Given the first wave is $y_1 = a \cos (kx + \omega t)$.
For a stationary wave to have a node at $x = 0$,the resultant displacement $y = y_1 + y_2$ must be zero at $x = 0$ for all $t$.
So,$y_1(0, t) + y_2(0, t) = 0$.
$a \cos (\omega t) + y_2(0, t) = 0$,which implies $y_2(0, t) = -a \cos (\omega t)$.
Since the waves must travel in opposite directions,if the first wave is traveling in the negative $x$-direction (as indicated by $kx + \omega t$),the second wave must travel in the positive $x$-direction,taking the form $y_2 = a' \cos (kx - \omega t + \phi)$.
At $x = 0$,$y_2(0, t) = a' \cos (- \omega t + \phi) = a' \cos (\omega t - \phi)$.
Comparing this with $-a \cos (\omega t)$,we get $a' = a$ and $\phi = \pi$.
Thus,$y_2 = a \cos (kx - \omega t + \pi) = -a \cos (kx - \omega t)$.

(B) The distance between two consecutive nodes in a stationary wave is given by $\frac{\lambda}{2}$.
Given the velocity $v = 20 \ m/s$ and frequency $f = n$.
The wavelength $\lambda$ is calculated as $\lambda = \frac{v}{f} = \frac{20}{n}$.
Therefore,the distance between two consecutive nodes is $\frac{\lambda}{2} = \frac{1}{2} \times \frac{20}{n} = \frac{10}{n}$.

(A) stationary wave (also known as a standing wave) is formed by the superposition of two identical waves traveling in opposite directions.
In a stationary wave,the energy is trapped between the nodes and does not propagate through the medium.
While the energy oscillates between kinetic and potential forms at different points,there is no net transport of energy across the medium.
In contrast,progressive waves,transverse waves,and electromagnetic waves are all characterized by the transport of energy from one point to another.
Therefore,the correct answer is $A$.

(C) The standard equation for a stationary wave is given by $y = A \cos(kx) \sin(\omega t)$.
Comparing the given equation $y = 5 \cos(\pi x / 3) \sin(40 \pi t)$ with the standard equation,we get the wave number $k = \pi / 3$.
We know that $k = 2\pi / \lambda$,where $\lambda$ is the wavelength.
Substituting the value of $k$: $\pi / 3 = 2\pi / \lambda$.
Solving for $\lambda$,we get $\lambda = 6 \, cm$.
The distance between two consecutive nodes in a stationary wave is given by $\lambda / 2$.
Therefore,the distance = $6 / 2 = 3 \, cm$.

(D) When two sinusoidal waves of the same frequency and amplitude travel in opposite directions,they form a stationary wave.
In a stationary wave,the string becomes flat (all particles pass through their mean position) twice in one time period $T$.
Therefore,the time interval between two consecutive instants when the string is flat is given by $\frac{T}{2}$.
Given,$\frac{T}{2} = 0.5 \, s$,which implies $T = 1.0 \, s$.
The relationship between wavelength $\lambda$,wave speed $v$,and time period $T$ is $\lambda = v \times T$.
Substituting the given values,$\lambda = 10 \, ms^{-1} \times 1.0 \, s = 10 \, m$.

(C) In a stationary wave (also known as a standing wave),the wave pattern does not propagate through the medium.
While individual particles of the medium perform $SHM$ with varying amplitudes,the wave itself does not transport energy from one point to another in the direction of the wave propagation.
Therefore,the net energy flow across any cross-section of the medium is zero.
This lack of energy transport is the defining characteristic that gives stationary waves their name.
Thus,the correct option is $C$.

(B) The distance between two consecutive nodes in a stationary wave is given by $\lambda/2$.
Given that the velocity of the waves is $v = 16 \ m/s$ and the frequency is $n$.
The relationship between velocity,frequency,and wavelength is $v = n\lambda$,which implies $\lambda = v/n$.
Substituting the value of $\lambda$ into the expression for the distance between two nodes:
Distance $= \lambda/2 = (v/n)/2 = v/(2n)$.
Substituting $v = 16 \ m/s$:
Distance $= 16/(2n) = 8/n$.

(D) Stationary waves are formed by the superposition of two waves of the same frequency and amplitude traveling in opposite directions.
Because the waves travel in opposite directions,the net energy transport across any cross-section is zero.
Stationary waves are characterized by fixed points of zero displacement called nodes and points of maximum displacement called antinodes.
Therefore,both statements $(b)$ and $(c)$ are correct.

(B) In a stationary wave,the medium is divided into segments by nodes (points of zero displacement).
All particles within a single segment (the region between two consecutive nodes) move in the same direction at any given instant and reach their extreme positions simultaneously.
Therefore,all particles in the region between two nodes vibrate in the same phase.

(A) stationary wave is formed by the superposition of two identical waves traveling in opposite directions.
If the incident wave is given by ${y_1} = a \sin(\omega t - kx)$ and the reflected wave is ${y_2} = a \sin(\omega t + kx)$,the resulting stationary wave is ${y} = {y_1} + {y_2} = 2a \sin(\omega t) \cos(kx)$.
In this expression,the angular frequency $\omega$ remains the same as that of the individual component waves.
Therefore,the frequency $f = \frac{\omega}{2\pi}$ of the stationary wave is the same as the frequency of the individual waves.

(B) In a stationary wave,a node is a point where the amplitude always remains zero at all times,and an antinode is a point where the amplitude is maximum at all times.
Since the energy of a wave is proportional to the square of its amplitude $(E \propto A^2)$,the energy is zero at the nodes (where $A = 0$) and maximum at the antinodes (where $A$ is maximum).
Therefore,energy is minimum at nodes and maximum at antinodes.

(A) The standard equation of a stationary wave is given by $y = A \sin(kx) \cos(\omega t)$.
Comparing the given equation $y = 10 \sin \left( \frac{\pi x}{4} \right) \cos(20 \pi t)$ with the standard form,we get the wave number $k = \frac{\pi}{4}$.
We know that $k = \frac{2\pi}{\lambda}$,where $\lambda$ is the wavelength.
Substituting the value of $k$: $\frac{2\pi}{\lambda} = \frac{\pi}{4}$.
Solving for $\lambda$: $\lambda = 8$.
The distance between two consecutive nodes in a stationary wave is given by $\frac{\lambda}{2}$.
Therefore,the distance $= \frac{8}{2} = 4$.

(A) In a longitudinal stationary wave,nodes are points where the displacement of particles is always zero.
Because the particles do not move at these points,the compression and rarefaction are most intense here.
Consequently,the change in pressure and density is maximum at the nodes.
Conversely,at antinodes,the displacement is maximum,but the pressure and density variations are minimum.

(A) standing wave (or stationary wave) is formed by the superposition of two waves of the same frequency and amplitude traveling in opposite directions along the same line.
Given:
$z_1 = A \sin(kx - \omega t)$ (traveling in $+x$ direction)
$z_2 = A \sin(kx + \omega t)$ (traveling in $-x$ direction)
$z_3 = A \sin(ky - \omega t)$ (traveling in $+y$ direction)
Adding $z_1$ and $z_2$:
$z_1 + z_2 = A \sin(kx - \omega t) + A \sin(kx + \omega t)$
Using the identity $\sin(C) + \sin(D) = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$z_1 + z_2 = 2A \sin(kx) \cos(\omega t)$
This represents a standing wave because the spatial part $\sin(kx)$ and temporal part $\cos(\omega t)$ are separated.
Waves $z_3$ and $z_1$ (or $z_2$) travel in different directions ($y$ and $x$ respectively),so they do not form a standing wave along a single axis.

(A) stationary wave is formed when two progressive waves of the same frequency and amplitude travel in opposite directions along the same line.
$1$. $Z_1 = A \cos(\omega t - kx)$ represents a wave traveling in the $+x$ direction with frequency $\omega$.
$2$. $Z_2 = A \cos(\omega t + kx)$ represents a wave traveling in the $-x$ direction with the same frequency $\omega$ and amplitude $A$.
$3$. Since $Z_1$ and $Z_2$ have the same frequency and travel in opposite directions along the same axis ($x$-axis),their superposition results in a stationary wave.
Therefore,the correct option is $A$.

(D) The superposition of two waves travelling in opposite directions with the same frequency and amplitude results in a stationary wave.
Using the principle of superposition,the resultant wave is $y = y_1 + y_2 = A\sin[k(x - ct)] + A\sin[k(x + ct)]$.
Using the trigonometric identity $\sin(C) + \sin(D) = 2\sin(\frac{C+D}{2})\cos(\frac{C-D}{2})$,we get:
$y = 2A\sin(kx)\cos(kct)$.
Nodes occur where the displacement is zero for all time $t$,which happens when $\sin(kx) = 0$.
This implies $kx = n\pi$,where $n = 0, 1, 2, \dots$.
The positions of the nodes are $x = \frac{n\pi}{k}$.
The distance between two adjacent nodes is the difference between consecutive positions:
$\Delta x = \frac{(n+1)\pi}{k} - \frac{n\pi}{k} = \frac{\pi}{k}$.
Alternatively,since $k = \frac{2\pi}{\lambda}$,the distance between adjacent nodes is $\frac{\lambda}{2} = \frac{\pi}{k}$.

(D) The given equation for the stationary wave is $y = 5 \sin \left( \frac{2\pi x}{3} \right) \cos (20\pi t)$.
Comparing this with the standard equation of a stationary wave $y = A \sin \left( \frac{2\pi x}{\lambda} \right) \cos (\omega t)$,we get $\frac{2\pi}{\lambda} = \frac{2\pi}{3}$.
Therefore,the wavelength $\lambda = 3 \ cm$.
The distance between two adjacent nodes in a stationary wave is given by $\frac{\lambda}{2}$.
Distance $= \frac{3}{2} = 1.5 \ cm$.

(C) The length of the string is $l = 10 \; m$. The number of segments is $p = 5$. The wave velocity is $v = 20 \; m/s$.
For a string vibrating in $p$ segments,the length $l$ is related to the wavelength $\lambda$ by the formula $l = p \cdot \frac{\lambda}{2}$.
Substituting the values,we get $10 = 5 \cdot \frac{\lambda}{2}$,which simplifies to $\lambda = \frac{10 \cdot 2}{5} = 4 \; m$.
The frequency $f$ is given by the relation $f = \frac{v}{\lambda}$.
Substituting the values,$f = \frac{20 \; m/s}{4 \; m} = 5 \; Hz$.

(C) In a standing wave,the distance between two consecutive nodes is equal to $\frac{\lambda}{2}$.
Given,the distance between nodes is $5.0 \ cm = 0.05 \ m$.
Therefore,$\frac{\lambda}{2} = 0.05 \ m$,which implies $\lambda = 0.1 \ m$.
The velocity of the wave is given as $v = 2 \ m/s$.
The frequency $f$ is calculated using the relation $v = f \lambda$.
Thus,$f = \frac{v}{\lambda} = \frac{2 \ m/s}{0.1 \ m} = 20 \ Hz$.

(C) For a string of length $l$ fixed at both ends,the boundary conditions require nodes at both ends.
This means the length $l$ must be an integer multiple of half-wavelengths: $l = n \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$ is the number of loops or the harmonic number.
Rearranging this equation for the wavelength $\lambda$,we get:
$\lambda = \frac{2l}{n}$.

(A) For a string fixed at both ends,the $n^{th}$ harmonic corresponds to $n$ loops.
In the $7^{th}$ harmonic,the string vibrates in $7$ loops.
Each loop consists of $1$ antinode,so the total number of antinodes is $7$.
The number of nodes in a string with $n$ loops is $n + 1$.
Therefore,the number of nodes $= 7 + 1 = 8$.
Thus,there are $8$ nodes and $7$ antinodes.