Stationary Waves (Standing wave) Questions in English

(B) For a string fixed at both ends,the $n^{th}$ overtone corresponds to the $(n+1)^{th}$ harmonic.
Here,the fifth overtone is the sixth harmonic $(n=6)$.
The length of the string $L = 2.4 \ m$.
The condition for the $n^{th}$ harmonic is $L = n \frac{\lambda}{2}$.
Substituting the values: $2.4 = 6 \times \frac{\lambda}{2}$.
This gives $\frac{\lambda}{2} = \frac{2.4}{6} = 0.4 \ m$.
Thus,$\lambda = 0.8 \ m$.
The distance between a node and the successive antinode is always $\frac{\lambda}{4}$.
Therefore,distance $= \frac{0.8 \ m}{4} = 0.2 \ m$.

(B) The resultant wave is formed by the superposition of two waves travelling in opposite directions: $Y_1 = A \sin(\omega t - kx)$ and $Y_2 = A \sin(\omega t + kx)$.
Using the trigonometric identity $\sin(C) + \sin(D) = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$,we get:
$Y = Y_1 + Y_2 = 2A \sin(\omega t) \cos(kx)$.
This represents a standing wave.
Nodes occur where the amplitude is zero,i.e.,$\cos(kx) = 0$.
This implies $kx = (2n+1) \frac{\pi}{2}$ for $n = 0, 1, 2, \ldots$.
Substituting $k = \frac{2\pi}{\lambda}$,we get $\frac{2\pi}{\lambda} x = (2n+1) \frac{\pi}{2}$.
Solving for $x$,we find $x = (n + \frac{1}{2}) \frac{\lambda}{2}$.

(C) The given progressive waves are $Y_{1} = \sin 2\pi(\frac{t}{0.4} - \frac{x}{4})$ and $Y_{2} = \sin 2\pi(\frac{t}{0.4} + \frac{x}{4})$.
Comparing these with the standard wave equation $Y = A \sin 2\pi(\frac{t}{T} \pm \frac{x}{\lambda})$,we get the wavelength $\lambda = 4 \ m$ and time period $T = 0.4 \ s$.
When two waves of equal amplitude and frequency traveling in opposite directions superpose,they form a standing wave given by $Y = Y_{1} + Y_{2} = 2A \cos(\frac{2\pi x}{\lambda}) \sin(\frac{2\pi t}{T})$.
Here,the amplitude of the standing wave at any position $x$ is given by $A_{res} = |2A \cos(\frac{2\pi x}{\lambda})|$.
Given $A = 1$,$\lambda = 4 \ m$,and $x = 0.5 \ m$,we substitute these values:
$A_{res} = 2 \times 1 \times |\cos(\frac{2\pi \times 0.5}{4})|$
$A_{res} = 2 \cos(\frac{\pi}{4})$
Since $\cos(45^{\circ}) = \frac{1}{\sqrt{2}}$,we get $A_{res} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} \ m$.

(B) When a string is fixed at both ends (rigid supports),the displacement at these points must be zero because they cannot move.
Points with zero displacement are called nodes.
Therefore,nodes are formed at both ends.
For the string to vibrate,there must be at least one point of maximum displacement between the two fixed ends,which is called an antinode.
Thus,the simplest mode of vibration (the fundamental mode) consists of nodes at both ends and at least one antinode in between.

(D) For a stretched string,the fundamental frequency (first harmonic) is $n = \frac{v}{2l} = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
The frequency of the second harmonic is $2n = \frac{1}{l} \sqrt{\frac{T}{m}}$.
The first overtone is the next possible frequency above the fundamental,which is $n_1 = \frac{2v}{2l} = \frac{v}{l} = \frac{1}{l} \sqrt{\frac{T}{m}}$.
Comparing these,the frequency of the first overtone is equal to the frequency of the second harmonic.
Therefore,option $D$ is correct.

(A) The velocity of the wave is given by $v = f \lambda$.
Given $v = 50 \,m/s$ and $f = 200 \,Hz$.
Calculating the wavelength $\lambda$:
$\lambda = \frac{v}{f} = \frac{50}{200} = 0.25 \,m$.
In a stationary wave, the distance between two consecutive antinodes is equal to half of the wavelength $(\frac{\lambda}{2})$.
Distance $= \frac{0.25 \,m}{2} = 0.125 \,m$.

(D) When two waves of the same frequency and velocity travel in opposite directions along the same path,they form a stationary wave.
Given frequency is $n$ and velocity $v = 12 \ m/s$.
The wavelength $\lambda$ is given by the relation $v = n\lambda$,so $\lambda = \frac{v}{n} = \frac{12}{n}$.
In a stationary wave,the distance between two consecutive nodes is equal to half the wavelength,i.e.,$d = \frac{\lambda}{2}$.
Substituting the value of $\lambda$,we get $d = \frac{1}{2} \times \frac{12}{n} = \frac{6}{n}$.

(C) The given equation for the stationary wave is $y = 12 \cos \left(\frac{\pi}{6} x\right) \sin (8 \pi t)$.
Comparing this with the standard form $y = A_0 \cos(kx) \sin(\omega t)$,we identify the wave number $k = \frac{2\pi}{\lambda}$.
From the equation,$k = \frac{\pi}{6}$.
Equating the two,we get $\frac{2\pi}{\lambda} = \frac{\pi}{6}$.
Solving for $\lambda$,we find $\lambda = 12 \ cm$.
The distance between two successive antinodes in a stationary wave is given by $\frac{\lambda}{2}$.
Therefore,the distance $= \frac{12 \ cm}{2} = 6 \ cm$.

(C) The length of the string is $L = 90 \ cm$.
There are $3$ nodes $(N)$ in the stationary wave.
The distance between two consecutive nodes is $\frac{\lambda}{2}$.
Since there are $3$ nodes,there are $2$ such segments (loops) between them.
Therefore,the total length $L = 2 \times \frac{\lambda}{2} = \lambda$.
Given $L = 90 \ cm$,we have $\lambda = 90 \ cm$.

(C) The resultant displacement $Y$ is given by the superposition principle: $Y = Y_1 + Y_2$.
Using the trigonometric identity $\sin(A - B) + \sin(A + B) = 2 \sin A \cos B$,where $A = \frac{2 \pi t}{0.4}$ and $B = \frac{2 \pi x}{4}$,we get:
$Y = 2 \sin \left( \frac{2 \pi t}{0.4} \right) \cos \left( \frac{2 \pi x}{4} \right)$.
The amplitude $R$ of the standing wave at any position $x$ is given by $R = |2 \cos \left( \frac{2 \pi x}{4} \right)|$.
Substituting $x = 0.5 \ m$:
$R = 2 \cos \left( \frac{2 \pi \times 0.5}{4} \right) = 2 \cos \left( \frac{\pi}{4} \right)$.
Since $\cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$,we have $R = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} \ m$.

(D) Stationary waves (also known as standing waves) are formed by the superposition of two identical waves traveling in opposite directions in a bounded medium.
These waves require a medium for propagation and reflection to occur at the boundaries.
Since mechanical waves can propagate through all three states of matter (solids,liquids,and gases),stationary waves can be produced in all these media.
For example,stationary waves are produced in a stretched string (solid),in a column of water (liquid),and in an air column (gas).

(A) The standard equation for a stationary wave is given by $y = 2a \sin(kx) \cos(\omega t)$,where $k = \frac{2\pi}{\lambda}$.
Comparing the given equation $Y = 10 \sin \left( \frac{\pi x}{15} \right) \cos (48 \pi t)$ with the standard form,we identify the wave number $k = \frac{\pi}{15} \text{ cm}^{-1}$.
Since $k = \frac{2\pi}{\lambda}$,we have $\frac{2\pi}{\lambda} = \frac{\pi}{15}$.
Solving for wavelength $\lambda$,we get $\lambda = 30 \text{ cm}$.
The distance between a node and the successive antinode in a stationary wave is given by $\frac{\lambda}{4}$.
Therefore,the distance $= \frac{30}{4} = 7.5 \text{ cm}$.

(C) In a stationary wave,the particles of the medium vibrate about their mean positions in $S.H.M.$
All particles (except those at the nodes,which remain at rest) vibrate with the same frequency (and thus the same period) as the source.
However,the amplitude of vibration varies from particle to particle,being zero at the nodes and maximum at the antinodes.

(D) The standard equation of a standing wave is $y = 2A \sin(kx) \cos(\omega t)$.
Comparing the given equation $Y = 0.5 \sin(0.314 x) \cos(600 \pi t)$ with the standard form,we identify the wave number $k = 0.314 \ cm^{-1}$.
Since $0.314 \approx \frac{\pi}{10}$,we have $k = \frac{\pi}{10} \ cm^{-1}$.
The wavelength $\lambda$ is given by $k = \frac{2\pi}{\lambda}$,so $\frac{\pi}{10} = \frac{2\pi}{\lambda}$,which gives $\lambda = 20 \ cm$.
For a string clamped at both ends,the length $L$ for the $n^{th}$ harmonic is $L = n \frac{\lambda}{2}$.
For the third harmonic,$n = 3$,so $L = 3 \times \frac{20}{2} = 3 \times 10 = 30 \ cm$.

(C) The given equation of the stationary wave is $y = 10 \sin \left( \frac{\pi x}{4} \right) \cos (20 \pi t)$.
Comparing this with the standard equation of a stationary wave,$y = 2A \sin (kx) \cos (\omega t)$,we get the propagation constant $k = \frac{\pi}{4} \ cm^{-1}$.
We know that the propagation constant $k$ is related to the wavelength $\lambda$ by the formula $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$: $\frac{\pi}{4} = \frac{2\pi}{\lambda}$.
Solving for $\lambda$,we get $\lambda = 8 \ cm$.
The distance between two consecutive nodes in a stationary wave is equal to half of the wavelength,which is $\frac{\lambda}{2}$.
Therefore,the distance is $\frac{8 \ cm}{2} = 4 \ cm$.

(C) stationary wave is formed by the superposition of two identical progressive waves traveling in opposite directions with the same frequency and amplitude.
In a stationary wave,the distance between two consecutive nodes or two consecutive antinodes is $\frac{\lambda}{2}$.
The wavelength of the stationary wave is defined as the distance between two consecutive points that are in the same phase,which is equal to the wavelength of the constituent progressive waves.
Therefore,the wavelength of the stationary wave is $\lambda$.

(C) In a stationary wave, the distance between two consecutive nodes is $\lambda / 2$.
For $4$ nodes, there are $3$ such segments of length $\lambda / 2$ between them.
Given the total length of the string is $L = 120 \,cm$.
Therefore, $3 \times (\lambda / 2) = 120 \,cm$.
$3 \lambda / 2 = 120 \,cm$.
$\lambda = (120 \times 2) / 3 \,cm$.
$\lambda = 240 / 3 \,cm$.
$\lambda = 80 \,cm$.

(A) In a stationary wave,the medium particles oscillate with different amplitudes at different positions.
Nodes are points where the amplitude of vibration is zero,meaning the displacement is always zero.
Antinodes are points where the amplitude of vibration is maximum.
Therefore,the displacement at a node is zero and at an antinode is maximum.

(A) The given equation is $Y = 0.04 \cos(\pi x) \sin(50 \pi t)$.
Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we can rewrite the equation as:
$Y = 0.02 \sin(50 \pi t + \pi x) + 0.02 \sin(50 \pi t - \pi x)$.
Comparing this with the standard wave equation $y = a \sin(\omega t \pm kx)$:
$1$. Amplitude $a = 0.02 \text{ m}$.
$2$. Angular frequency $\omega = 50 \pi \text{ rad/s}$.
$3$. Wave number $k = \pi \text{ m}^{-1}$.
Now,calculating the parameters:
- Time Period $T = \frac{2 \pi}{\omega} = \frac{2 \pi}{50 \pi} = 0.04 \text{ s}$.
- Wavelength $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{\pi} = 2 \text{ m}$.
- Velocity $v = \frac{\omega}{k} = \frac{50 \pi}{\pi} = 50 \text{ m/s}$.
Comparing these results with the given options,the statement in option $A$ (Time Period $= 0.02 \text{ s}$) is incorrect,as the calculated time period is $0.04 \text{ s}$.

(A) The standard equation for a stationary wave is given by $y = A \sin(kx) \cos(\omega t)$.
Comparing the given equation $y = 4 \sin \left(\frac{\pi x}{15}\right) \cos (96 \pi t)$ with the standard form,we identify the wave number $k$ as:
$k = \frac{2\pi}{\lambda} = \frac{\pi}{15}$
Solving for the wavelength $\lambda$:
$\lambda = 15 \times 2 = 30$
The distance between a node and the adjacent antinode in a stationary wave is always $\frac{\lambda}{4}$.
Therefore,the required distance is $\frac{30}{4} = 7.5$.

(B) When two waves of the same frequency and velocity travel in opposite directions,they form a standing wave.
The distance between two consecutive nodes in a standing wave is given by $x = \frac{\lambda}{2}$.
We know that the wave speed $v$ is related to frequency $n$ and wavelength $\lambda$ by the formula $v = n \lambda$,which implies $\lambda = \frac{v}{n}$.
Substituting the value of $\lambda$ into the distance formula:
$x = \frac{1}{2} \left( \frac{v}{n} \right) = \frac{v}{2n}$.
Given $v = 20 \ m/s$,we have:
$x = \frac{20 \ m/s}{2n} = \frac{10}{n} \ m$.

(B) Given:
Velocity of waves,$v = 50 \,m/s$
Frequency of waves,$f = 200 \,Hz$
First,we calculate the wavelength $\lambda$ of the progressive waves:
$v = f \lambda$
$\lambda = \frac{v}{f} = \frac{50}{200} = 0.25 \,m$
When two identical progressive waves travel in opposite directions,they form a standing wave.
In a standing wave,the distance between two consecutive antinodes is equal to half the wavelength $(\frac{\lambda}{2})$.
Distance $= \frac{\lambda}{2} = \frac{0.25 \,m}{2} = 0.125 \,m$
Therefore,the correct option is $B$.

(C) Given: Linear mass density $\mu = 0.1 \ kg/m$,Length $L = 0.9 \ m$,Tension $T = 40 \ N$,Amplitude $A = 0.3 \ cm = 0.003 \ m$,Number of segments $n = 3$.
Wave speed $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{40}{0.1}} = \sqrt{400} = 20 \ m/s$.
The frequency of the $n^{th}$ harmonic is $f_n = \frac{n v}{2L} = \frac{3 \times 20}{2 \times 0.9} = \frac{60}{1.8} = \frac{600}{18} = \frac{100}{3} \ Hz$.
Angular frequency $\omega = 2 \pi f_n = 2 \pi \times \frac{100}{3} = \frac{200 \pi}{3} \ rad/s$.
Maximum particle velocity $v_{max} = A \omega = 0.003 \times \frac{200 \pi}{3} = \frac{3}{1000} \times \frac{200 \pi}{3} = \frac{\pi}{5} \ m/s$.

(C) For a string fixed at both ends,the wave forms a stationary wave pattern.
Let $L$ be the length of the string and $\lambda$ be the wavelength.
The number of loops formed in the string is $n$,where $n = x - 1$ (since there are $x$ nodes including the two fixed ends).
The length of the string is given by $L = n \frac{\lambda}{2}$.
Substituting $n = x - 1$,we get $L = (x - 1) \frac{\lambda}{2}$.

(B) When a string is fixed at both ends,waves travel along the string and reflect back from the fixed boundaries. The superposition of the incident and reflected waves results in the formation of stationary (standing) waves. Since the displacement of the string particles is perpendicular to the direction of wave propagation,these waves are transverse in nature. Therefore,the waves are stationary transverse waves.

(D) In a standing wave on a string fixed at both ends,the string is divided into segments called loops.
Each loop is separated by nodes (points of zero displacement).
Within a single loop,all particles vibrate in phase with each other.
Particles in adjacent loops are separated by a node and vibrate with a phase difference of $\pi$ radians (i.e.,they are in anti-phase).
Therefore,particles in alternate loops (separated by an even number of nodes) vibrate in phase.
Since each antinode is located at the center of a loop,the particles at all antinodes vibrate in phase with each other.

(B) The given equation of the stationary wave is $ y = 2 \sin \left( \frac{\pi x}{15} \right) \cos (48 \pi t) $.
Comparing this with the standard equation $ y = A \sin(kx) \cos(\omega t) $, we get the wave number $ k = \frac{\pi}{15} $.
We know that $ k = \frac{2 \pi}{\lambda} $, so $ \frac{2 \pi}{\lambda} = \frac{\pi}{15} $.
Solving for wavelength $ \lambda $, we get $ \lambda = 30 $ units.
The distance between a node and its consecutive antinode is given by $ d = \frac{\lambda}{4} $.
Substituting the value of $ \lambda $, we get $ d = \frac{30}{4} = 7.5 $ units.

(D) The general equation of a stationary wave is given by $y = A \sin(kx) \cos(\omega t)$.
Comparing this with the given equation $y = 0.7 \sin \left(\frac{7 \pi}{4} x\right) \cos (350 \pi t)$,we get:
Wave number $k = \frac{7 \pi}{4} \ m^{-1}$
Angular frequency $\omega = 350 \pi \ rad \ s^{-1}$
The wave speed $v$ is defined as the ratio of angular frequency to wave number:
$v = \frac{\omega}{k}$
Substituting the values:
$v = \frac{350 \pi}{7 \pi / 4}$
$v = 350 \pi \times \frac{4}{7 \pi}$
$v = 50 \times 4 = 200 \ m \ s^{-1}$
Thus,the speed of the wave is $200 \ m \ s^{-1}$.

(A) In a stationary wave formed on a string fixed at both ends,the points where the displacement is always zero are called nodes. The points where the amplitude of vibration is maximum are called antinodes.
For a string vibrating in $n$ loops,the number of nodes is $n + 1$ and the number of antinodes is $n$.
Given that the string vibrates in $5$ loops,we have $n = 5$.
Therefore,the number of nodes = $5 + 1 = 6$.
The number of antinodes = $5$.
Thus,the total number of nodes and antinodes are $6$ and $5$ respectively.

(B) Frequency, $f = 1000 \,Hz$.
Total distance between $6$ successive nodes is $d = 85 \,cm = 0.85 \,m$.
In a stationary wave, the distance between two successive nodes is $\frac{\lambda}{2}$.
Therefore, the distance between $6$ successive nodes (which involves $5$ intervals of $\frac{\lambda}{2}$) is given by:
$d = 5 \times \frac{\lambda}{2} = \frac{5 \lambda}{2}$.
Equating this to the given distance:
$\frac{5 \lambda}{2} = 0.85 \,m$.
$\lambda = \frac{0.85 \times 2}{5} = 0.34 \,m$.
The speed of sound $v$ is given by $v = f \lambda$.
$v = 1000 \,Hz \times 0.34 \,m = 340 \,ms^{-1}$.

(C) In a stationary wave,nodes $(N)$ are points of minimum displacement (zero amplitude),and anti-nodes $(A)$ are points of maximum displacement (maximum amplitude).
The distance between two consecutive nodes is $\frac{\lambda}{2}$.
The distance between two consecutive anti-nodes is $\frac{\lambda}{2}$.
The distance between a node and the next consecutive anti-node is exactly half the distance between two consecutive nodes,which is $\frac{1}{2} \times \frac{\lambda}{2} = \frac{\lambda}{4}$.
Therefore,the distance between the successive node and anti-node is $\frac{\lambda}{4}$.

(A) In a standing wave,the distance between two consecutive nodes is $\frac{\lambda}{2}$.
Given that there are $3$ nodes and $2$ antinodes between two atoms,the total distance $L$ between the two atoms corresponds to two segments of length $\frac{\lambda}{2}$ each.
Therefore,the total distance $L = \frac{\lambda}{2} + \frac{\lambda}{2} = \lambda$.
Given $L = 1.21 \ Å$,we have $\lambda = 1.21 \ Å$.

(C) The given equation of the stationary wave is $y = 20 \sin(\pi x) \cos(\omega t)$.
Comparing this with the standard equation $y = A \sin(kx) \cos(\omega t)$,we get the wave number $k = \pi \text{ rad/m}$.
We know that $k = \frac{2\pi}{\lambda}$,so $\frac{2\pi}{\lambda} = \pi$,which gives the wavelength $\lambda = 2 \text{ m} = 200 \text{ cm}$.
The distance between a node and its adjacent antinode is given by $\frac{\lambda}{4}$.
Therefore,the distance $= \frac{200 \text{ cm}}{4} = 50 \text{ cm}$.

(D) The given equation for the standing wave is $y_{(x, t)} = 0.03 \sin \left(\frac{2 \pi x}{3}\right) \cos (60 \pi t)$.
Comparing this with the standard standing wave equation $y = a \sin(kx) \cos(\omega t)$,we get:
Angular frequency $\omega = 60 \pi \text{ rad s}^{-1}$.
Wave number $k = \frac{2 \pi}{3} \text{ m}^{-1}$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{60 \pi}{2 \pi / 3} = 30 \times 3 = 90 \text{ m s}^{-1}$.
The speed of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Given $\mu = 0.01 \text{ kg m}^{-1}$.
Substituting the values: $90 = \sqrt{\frac{T}{0.01}}$.
Squaring both sides: $8100 = \frac{T}{0.01}$.
$T = 8100 \times 0.01 = 81 \text{ N}$.

(B) Given,length of the string,$L = 16 \ m$.
Speed of the wave,$v = 32 \ m/s$.
Number of nodes between the two fixed ends $= 9$.
The total number of nodes including the fixed ends is $9 + 2 = 11$.
The number of loops or segments $(p)$ formed in the string is given by the number of nodes minus $1$,so $p = 11 - 1 = 10$.
The frequency of the $p^{th}$ harmonic is given by the formula $f = \frac{p \cdot v}{2L}$.
Substituting the values: $f = \frac{10 \times 32}{2 \times 16}$.
$f = \frac{320}{32} = 10 \ Hz$.

(B) The standard equation for a stationary wave is $y = A \sin(kx) \cos(\omega t)$.
Comparing this with the given equation $y = 5 \sin \left( \frac{\pi x}{3} \right) \cos(40 \pi t)$,we identify the wave number $k = \frac{\pi}{3} \ cm^{-1}$.
The relationship between the wave number $k$ and the wavelength $\lambda$ is $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$: $\frac{\pi}{3} = \frac{2\pi}{\lambda}$,which gives $\lambda = 6 \ cm$.
The separation between two adjacent nodes in a stationary wave is given by $\frac{\lambda}{2}$.
Therefore,the separation $= \frac{6 \ cm}{2} = 3 \ cm$.

(A, B, D) The given equation is $y(x, t) = 0.02 \cos(50 \pi t + \frac{\pi}{2}) \cos(10 \pi x)$.
Nodes occur where the spatial part $\cos(10 \pi x) = 0$.
$10 \pi x = (n + \frac{1}{2}) \pi \implies x = \frac{n + 0.5}{10} = 0.05, 0.15, 0.25, \dots \ m$. Thus,a node occurs at $x = 0.15 \ m$.
Antinodes occur where $|\cos(10 \pi x)| = 1$.
$10 \pi x = n \pi \implies x = \frac{n}{10} = 0, 0.1, 0.2, 0.3, \dots \ m$. Thus,an antinode occurs at $x = 0.3 \ m$.
Comparing with the standard stationary wave equation $y = A \cos(\omega t + \phi) \cos(kx)$,we get $k = 10 \pi$ and $\omega = 50 \pi$.
Wavelength $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{10 \pi} = 0.2 \ m$.
Wave speed $v = \frac{\omega}{k} = \frac{50 \pi}{10 \pi} = 5 \ m/s$.