Waves PDF

# Waves ## Wave Motion - A wave motion is a way of transferring energy or information. - A wave forms when a system is disturbed from its equilibrium position, and this disturbance travels from one region of the system to another. - Thus, in a wave motion, the disturbance travels through the same...

# Waves ## Wave Motion - A wave motion is a way of transferring energy or information. - A wave forms when a system is disturbed from its equilibrium position, and this disturbance travels from one region of the system to another. - Thus, in a wave motion, the disturbance travels through the same medium, but the medium does not travel along with the disturbance. ## Types of Waves - There are 3 types of waves: - Mechanical waves - Electromagnetic waves - Matter waves ### Mechanical Waves - Mechanical waves can be produced only in a material medium such as water and air. - Examples: - Waves on water surface - Waves on strings of sitar - Sound waves ### Electromagnetic waves - Electromagnetic waves require no material medium for their propagation. - They can travel through vacuum and any other material medium. - Examples: - Visible light - UV light - Radio waves ### Matter waves - Matter waves are associated with moving particles of matters like electrons, protons, neutrons, etc. ## Types of Mechanical Wave Motion - Transverse wave motion: - Transverse wave motion is a motion in which particles of the medium execute simple harmonic motion (SHM) about their mean positions in a direction perpendicular to the direction of propagation of the wave motion. - Example: Movement of a string of a sitar. - Longitudinal wave motion: - A longitudinal wave motion is a motion in which individual particles of the medium execute SHM about their mean positions along the same direction along which the wave is propagated. - Example: Movement of a kink on a rope. #### Compression - A compression is a region of the medium in which particles come closer. #### Rarefaction - A rarefaction is a region of the medium in which particles are rarefied, i.e., particles get farther apart than they normally are. ## Speed of a Traveling Wave - Suppose a transverse wave is traveling on a string in the positive direction of the x-axis. - A wave by solid line is transverse waves at t=0 and same transverse waves at t=t_ot are shown by dotted line. - The equation of the wave is: $$y(x, t) = r sin(wt-kx)$$ - For points to retain their displacement φ must be constant: $$φ = wt-kx = Constant, if t = t₀ x must also increase$$ - To keep an argument constant, differentiate equation 1 with respect to time: $$d(wt-kx) = 0$$ $$ \frac {d(wt-kx)}{dt} = 0 $$ $$w - k\frac {dx}{dt} = 0$$ $$w = k\frac {dx}{dt} = kv$$ $$w = kv \rightarrow v = \frac {w}{k}$$ - $$w= 2πf$$ - $$k = \frac{2π}{λ}$$ $$v = \frac {w}{k} = \frac{2πf}{\frac{2π}{λ}} = fλ$$ $$v = \frac {w}{k} = \frac{2πf}{\frac{2π}{λ}} = λf$$ $$ v = uλ $$ ## Speed of Transverse Waves on a Stretched String - Speed of transverse waves (v) on a string depends on: - Linear mass density (µ), i.e., mass per unit length. - Tension (T) in the string. - Let $$ v = µ^a T^b $$ $$ v = km^aT^b $$ - Where: - v is the speed of the wave. - k is a constant. - m is mass. - T is tension. - Units: - v = [LT^-1], m = [M], T = [MLT^-2]. - Substituting the units in the equation: $$[LT^-1] = [M]^a[MLT^-2]^b$$ $$ [MLT^-1] = M^a+bL^bT^-2b $$ - Comparing the powers: - a+b = 0, b = 1, -2b = -1. - Solving for a and b, we get: - a = -1/2, b = 1/2. - Putting the values of a and b in the equation: $$v = km^-1/2T^1/2$$ $$ v= k \sqrt{\frac{T}{m}} $$ - Speed does not depend upon the frequency of the wave. - Speed of transverse waves in a solid is given by: $$ v= \sqrt{\frac{E}{ρ}} $$ - Where E is Young's modulus and p is density. ## Speed of Longitudinal Waves <start_of_image> え - Speed of longitudinal wave depends upon: - Bulk modulus (B) - Density (ρ) of the medium. - $$ v = \sqrt{\frac{B}{ρ}} $$ - $$ v = kρ^aB^b $$ - Where: - v is the speed of the wave. - k is a constant. - ρ is density. - B is the bulk modulus. - Units: - v = [LT^-1], p = [ML^-3], B = [ML^-1T^-2]. - Substituting the units in the equation: - $$ [MLT^-1] = [ML^⁻3]^a[ML^⁻1T^-2]^b $$ - $$ [MLT^-1] = M^a+bL^-3a-bT^-2b $$ - Comparing the powers: - a+b = 1, -3a-b = 1, -2b = -1. - Solving for a and b, we get: - a = -1/2, b = 1/2. - Putting the values of a and b in the equation: - $$ v = kρ^-1/2B^1/2$$ $$v = k \sqrt{\frac {B}{ρ}} $$. - This is the expression for the speed of a longitudinal wave in a fluid. ## Newton's Formula for Velocity of Sound in Gases - As sound travels through a gas in the form of compressions and rarefactions, it’s assumed that change in pressure and volume of a gas when waves are propagated through it are isothermal. - The amount of heat produced during compression is lost to the surroundings, and similarly, the amount of heat lost during rarefaction is gained from the surroundings, so as to keep the temperature constant. - $$ v= \sqrt{\frac{B_{i}}{ρ}} $$ - where B_i is the isothermal Bulk modulus. - Calculating B_i; - Let: - P = Initial pressure of the gas. - V = is the initial volume of the gas. - Using isothermal conditions PV = Constant. - $$ P dV + v dP = 0$$ - $$PdV = -v dP$$ - $$P= -v \frac{dP}{dV} = B_i $$ - Now, - $$v= \sqrt{\frac{B_{i}}{ρ}} = \sqrt{\frac {P}{ρ}}$$. - Solving the error in Newton’s formula; - $$v= \sqrt{\frac {P}{ρ}} $$ - $$v= \sqrt{\frac {g'}{g}} \sqrt{\frac {P}{ρ}}$$ - Where: - g' = Density of Hg = 13.6 x 10³ kgm^-3. - g = 9.8 ms^-2. - ρ = Density of air = 1.293kgm^-3. - Then, $$v=0.76 × 13.6 × 10^{ 3} ×1.8$$ $$v= 280 ms^-1$$ - Experimental value of velocity of sound in air at NTP is 332 ms^-1. - Error = Experimental value – Measured value - $$Error= 332 – 280 = 52 ms^-1$$ - Percentage Error = 52/332 × 100 = 15.7/ 332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 = 15.7/332 ≈ 5%. - Thus, the value calculated on the basis of Newton’s formula was less than the experimental value of the velocity of sound in the air by about 16%. Such a large error could not be taken as an experimental error. ## Laplace Correction - Laplace pointed out that Newton's assumption was wrong. - According to Laplace, the change in pressure and volume of gas when sound waves are propagated through it are not isothermal but adiabatic. - Because the velocity of sound in a gas is quite large, pulses of compression and rarefaction follow one another so rapidly that there is no time left for any exchange of heat. - Thus, no exchange of heat is possible when sound waves pass through a gas. - Hence, the changes are adiabatic and not isothermal. - $$ v= \sqrt{\frac{B_{a}}{ρ}} $$ - B_ais the adiabatic bulk modulus. - Calculation of B_a: - For adiabatic conditions: PV^Y = Constant. - $$P(v^Y-dV^Y) + v^YdP = 0$$ - $$YPV^Y-1dV = -v^YdP$$ - $$ YP dV = - v dP= B_a $$ - $$YP = -\frac{dP}{dV/V} = B_a $$ - Corrected formula: - $$ v = \sqrt{\frac{B_{a}}{ρ}} = \sqrt{\frac{Y P}{ρ}} $$. - where, Y = c_p/c_v = 1.41. - $$ v = \sqrt{Y} \sqrt{\frac{P}{ρ}} $$. - Then, - $$ v= \sqrt{1.41} × \sqrt{280}$$ - $$ v= 332.5 ms^-1 $$ - This value agrees well with the experimental value of velocity of sound in air at NTP. ## Factors Affecting Velocity of Sound - (a) **Effect of density:** - Velocity of sound in a gaseous medium: $$ v= \sqrt{\frac{γP}{ρ}} $$ - Velocity of sound in a gas is inversely proportional to the square root of density of the gas. - Example: The density of oxygen is 16 times the density of hydrogen. $$ v_{o}= \sqrt{\frac{γP}{ρ_{o}}} = \sqrt{\frac{γP}{16ρ_{H}}} = \frac{ 1} {4} $$ $$ v_{o} = \frac {1}{4} v_{H}$$ $$ v_{H} = 4v_{o} $$ - Therefore, the velocity of sound in hydrogen is 4 times greater than that in oxygen. - (b) **Effect of pressure:** - $$v = \sqrt{\frac{γP}{ρ}} $$ - $$v = \sqrt{\frac{γM}{V}} $$ - for constant T, PV = Constant. - Velocity of sound is independent of the change in pressure of the gas (provided T is constant). - (c) **Effect of temperature:** - We know, $$ v = \sqrt{\frac{γP}{ρ}} $$. - $$PV= RT$$ - $$P = \frac {RT}{V}$$ - $$v= \sqrt{\frac{γRT}{V}}, v = \sqrt{\frac{γRT}{M}}$$ - $$ v = \sqrt{γRT} $$, v is directly proportional to the square root of its absolute temperature. - **Temperature Coefficient of Velocity of Sound in Air (α):** - Temperature coefficient of velocity of sound in air is the change in velocity of sound in air when the temperature changes by 1°C. - $$v_t = velocity of sound in air at t°C$$ - $$v_o = velocity of sound in air at 0°C$$ - $$α = \frac{v_t - v_o}{t}$$ - Unit of α = m/s °C-1 - $$ \frac{v_t}{v_o} = \sqrt{\frac{t+273}{0+273}} = \sqrt{\frac{(t+273)}{ 273}}$$. - $$ v_t = (1+\frac{t}{273})^{\frac{1}{2}} v_o $$ - $$ v_t = (1 + \frac{t}{273}) v_o $$ - $$ v_t - v_o = \frac{t}{546} v_o $$ - $$ v_t - v_o = αt $$ - $$ α = \frac {v_t - v_o} {t}=\frac {v_o} {546}$$ - taking $v_{o} = 332 ms^{-1}$: - $$ α = \frac{332}{546} = 0. 608 ms^{-1} °C^{-1 } $$ - Therefore, the velocity of sound in air increases by 0.61 m/s^-1 for every 1°C rise in temperature. - (d) **Effect of Humidity:** - $$ v = \sqrt{\frac{γP}{ρ}} $$ - $$ v_m= \sqrt{\frac{γP}{ρ_{m}}} $$ - Where: - $$ρ_m = density of moist air$$ - $$ρ_d = density of dry air$$ - $$ v_m = Velocity of sound in moist air$$ - $$ v_d = Velocity of sound in dry air$$ - $$ v_m = \sqrt{\frac{ρ_d} {ρ_{m}}} v_d $$ - The presence of water vapor reduces the density of air: $$ρ_d > ρ_{m}$$ - $$v_m > v_d$$ - Therefore, the velocity of sound in moist air is greater than in dry air. - (e) **Effect of wind velocity:** - Wind velocity (W) can be resolved into 2 rectangular components: - W cos θ along SL - W sin θ along L to V - Component W sin θ has no effect on V. - Since v and W cos θ act in the same direction. - Resultant velocity of sound along SL $$ = v + w cos θ$$ ## Difference between Transverse Waves and Longitudinal Waves | Transverse Waves | Longitudinal Waves | |------------------------------------------------------------------------------------|----------------------------------------------------------------------------------------------------| | Particle motion is perpendicular to the direction of wave propagation. | Particle motion is parallel to the direction of wave propagation. | | Can propagate through solids, liquids and vacuum. | Typically require a medium (solid, liquid or gas) for propagation. | | Examples: light waves, waves on a string, waves on a water surface. | Examples: Sound waves in air. | | Speed depends on tension and mass per unit length in a medium (for mechanical waves) | Speed depends on density & elasticity of medium. | ## Reflection of Waves - Reflection of waves at an open boundary or an open end / Rigid boundary - Suppose incident wave is represented by: - $$ y(x, t) = r sin (wt-kx) $$ - On reflection at a rigid boundary, the direction of motion is reversed, and a phase change of π is introduced: - $$y(x, t) = r sin (wt + kx + π) = -r sin (wt + kx)$$ - Reflected wave travels along the negative direction of the x-axis. So, we are using t+kx. - On reflection at an open boundary, no phase change occurs, though the direction of motion is reversed. - Therefore, the reflected wave is represented by: - $$y(x, t) = r sin (wt + kx)$$ ## Principle of Superposition of Waves - When any number of waves meet simultaneously at any point in a medium, the net displacement at the given point is the sum of displacements due to each wave at the given point at the same time. Let the 1st wave be: $$ y₁(x, t) = r sin(wt-kx)$$ Let the 2nd wave be: $$ y₂ (x, t) = r sin(wt-kx+φ)$$ According to the superposition principle: - $$ y(x, t) = y₁(x, t)+ y₂(x, t) $$ - $$y(x, t) = r sin(wt-kx)+ r sin(wt-kx + φ)$$ - $$y = 2r sin(\frac {wt-kx+ wt-kx + φ} {2} ) cos (\frac{ wt-kx-wt-kx-φ} {2}) $$ - $$y = 2r sin(wt-kx+\frac{φ}{2}) cos (\frac{-φ}{2})$$ - Since sin(-D) = -sin(D) and cos(-D) = cos(D), - $$y = 2r sin(wt-kx+\frac{φ}{2}) cos (\frac{φ}{2})$$ - $$y = 2r cos(\frac {φ}{2}) sin(wt-kx+\frac{φ}{2})$$ - Three important applications of the superposition principle are: - Stationary waves. - Beats. - Interference of waves. ## Standing Waves or Stationary Waves - When two progressive waves of the same type (both longitudinal or both transverse) having the same amplitude, the same time period, the same wavelength, and travelling with the same speed along the same straight line in opposite directions, superimpose, a new set of waves are formed. These are known as stationary waves or standing waves. ## Characteristics of Standing Waves or Stationary Waves 1. The disturbance is confined to a particular region between the starting point and the reflecting point of the wave. 2. There is no onward motion of the disturbance from one particle to the adjoining particle, and so on beyond this particular region. 3. The total energy associated with a stationary wave is twice the energy of each of the incident and the reflected wave. 4. There are certain points in the medium in standing waves which are permanently at rest. These are known as nodes. - Distance between two consecutive nodes is λ/2. 5. The point of maximum amplitude of vibration in a standing wave are known as antinodes. - Distance between two consecutive nodes is λ/2 and distance between two consecutive antinodes is λ/2. - Distance between a node and an adjoining antinode is λ/4. 6. All particles except those at nodes execute simple harmonic motion about their mean position. 7. The amplitude of vibration of particles varies from zero at nodes to maximums at antinodes. ## Standing Waves in Strings & Normal Modes of Vibration: Analytical treatment - A wave pulse on a string (from left to right) in the direction of the x-axis is: $$ y₁(x, t) = r sin(wt - kx) $$ - For the reflected wave, the phase change of π on reflection at the fixed end B is: - $$ y₂(x, t) = r sin(wt + kx + π) = -r sin(wt + kx) $$ - For the wave pulse traveling from right to left, - $$ y₃(x, t) = - r sin(wt + kx) $$ - According to the superposition principle, the resultant displacement is: $$ y(x, t) = y₁(x, t) + y₂(x, t) $$ $$ y(x, t) = r sin(wt - kx) - r sin(wt + kx) $$ $$ y(x, t) = r [sin(wt - kx) - sin(wt + kx)] $$ $$ y(x, t) = 2r sin (\frac{wt - kx + wt + kx}{2}) cos(\frac{wt - kx - wt - kx}{2}) $$ - $$y(x, t) = 2r cos(kx) sin(wt) $$ - $$y(x, t) = -2r sin(kx) cos(wt) $$ - At one end of a string where x = 0, - $$ y = -2r sin(kL) cos(wt)= 0 $$. This end is the node. - At the other end of the string x = L, $$ y = -2r sin(kL) cos(wt) $$ - $$ y = 0, sin (kL) = 0 $$ - $$ sin(KL) = sin(nπ) $$ - $$ KL = nπ $$ - $$ k = \frac{nπ}{L} $$ - $$ \frac{2π}{λ} = \frac{nπ}{L} $$ - $$ λ = \frac{2L}{n} $$ - n = 1, 2, 3 corresponds to 1st, 2nd, and 3rd normal modes of vibration. - λ = 2L, λ₁= 2L or L = λ/2 , λ₂= 2L/2 or L = λ. ## Frequency of Vibration - $$ D_n = \frac{v}{λ} = \frac {v} {2L/n} = \frac{nv}{2L} = n(\frac{v}{2L}) $$ - This (first) normal mode of vibration is known as the fundamental mode. - The frequency of vibration of the string in this mode is minimum, and it is also known as fundamental frequency. - A sound or note so produced is known as the fundamental note or first harmonic. ## Second Normal Mode of Vibration - $$ λ = λ₂ = \frac{2L}{2} = L$$ - $$ D₂ = \frac{v}{λ} = \frac {v} {2L/2} = \frac{2v}{2L} = 2(\frac{v}{2L}) $$ - $$ D₂ = 2D₁ $$ - The frequency of vibration of a string becomes twice the fundamental frequency. - The note so produced is known as the second harmonic or first overtone. ## Third Normal Mode of Vibration - $$ λ = λ₃= \frac{2L}{3}$$ - $$ D₃ = \frac{v}{λ} = \frac {v} {2L/3} = \frac{3v}{2L} = 3(\frac{v}{2L})$$ - $$ D₃ =3D₁ $$ - The frequency of vibration of a string becomes 3 times the fundamental frequency. - This note is known as the third harmonic or second overtone. ## General - $$D_n = \frac{ν}{λ} = \frac{v}{2L/n} = n(\frac{v}{2L}) $$ - $$ D_n=nD₁ $$ --- This frequency is “n” times the fundamental frequency. So, the nth harmonic (the nth normal mode) is produced with this frequency. ## Position of Nodes - x = 0, $\frac{L}{2n+1}$ , $\frac{3L}{2n+1}$ ,$\frac{5L}{2n+1}$ ,........$\frac{(2n+1)L}{2n+1}$ - Example: In the first normal mode (n = 1) the nodes are at x = 0, x = L. - In the second normal mode (n = 2) the nodes are at x = 0, x = (1/3)L, x = (2/3)L. ## Positions of Antinodes - x = $\frac{L}{2n+1}$, $\frac{3L}{2n+1}$, $\frac{5L}{2n+1}$ , ........ $\frac{(2n+1)L}{2n+1}$. - Example: In the first normal mode (n = 1) the antinode is at x = L/2. - In the second normal mode (n = 2) the antinodes are at x = L/4, x = 3L/4. ## Standing Waves in Closed Organ Pipes: Analytical Treatment - An organ pipe is a wind instrument which is used for producing musical sound by setting the air column inside the pipe into vibration. - Examples of organ pipes: - Trumpet - Flute - Shehnai ## Organ Pipe: - Closed organ pipe: <start_of_image> - Open organ Pipe: <start_of_image> - **For closed organ pipe:** - According to superposition: - $$ y₁ (x, t) = r sin(wt-kx) $$ - $$y₂(x, t) = r sin(wt+kx+π) = -r sin(wt+kx)$$ - $$y = y₁ + y₂$$ - $$y(x, t) = r sin(wt-kx) - r sin(wt+kx) $$ - $$y(x, t) = r [sin(wt-kx) - sin(wt + kx)] $$ - $$y(x, t) = 2r sin (\frac{wt - kx + wt + kx}{2}) cos(\frac{wt - kx - wt - kx}{2}) $$ - $$y(x, t) = -2r cos(kx) sin(wt) $$ - $$y(x, t) = -2r sin(kx) cos(wt) $$ - At the closed end of the pipe, x = 0 - $$y = -2r sin(kL) cos(wt) = 0 $$ - A node is formed at the closed end. - At the open end of the pipe of length L, - $$y = L, an antinode is being formed $$ - $$ sin(KL) = max = ±1 = sin(2n + 1)\frac{π}{2}$$ - Where n = 1, 2, 3, ……… - $$KL = (2n + 1)\frac{π}{2} $$ - $$Kλ = (2n + 1) \frac{π}{2} $$ - $$\frac{2π}{λ}λ = (2n + 1) \frac{π}{2} $$ - $$\frac{4L}{λ} = (2n + 1) $$ - $$λ = \frac{4L}{(2n + 1)} $$ - The phase of y₂ reverses on reflection. - (i) **First normal mode of vibration:** - n = 1 - $$ λ= λ₁ = \frac{4L}{2x 1+1} = \frac{4L}{3} $$ - $$ D₁ = \frac{v}{λ} = \frac {v} {4L/3} = \frac{3v}{4L} $$ - $$D₁= \frac{V}{4L} $$ - This is known as the fundamental frequency. - The note so produced is known as the fundamental note or first harmonic. - (ii) **Second normal mode of vibration:** - n = 2 - $$ λ= λ₂ = \frac{4L}{2x2+1} = \frac{4L}{5} $$ - $$ D₂ = \frac{v}{λ} = \frac {v} {4L/5} = \frac{5v}{4L} $$ - $$D₂= 3D₁ = 3(\frac{v}{4L}) = 3D₁ $$ - $$D₂= 3D₁ $$ - The note so produced is known as the 3rd harmonic or first overtone. - (iii) **Third normal mode of vibration:** - n = 3 - $$ λ= λ₃= \frac{4L}{2x3+1} = \frac{4L}{7} $$ - $$ D₃ = \frac{v}{λ} = \frac {v} {4L/7} = \frac{7v}{4L} $$ - $$D₃= 5D₁ = 5(\frac{v}{4L}) = 5D₁ $$ - $$D₃= 5D₁ $$ - This is the 5th Harmonic or second overtone. - **Note:** - In general, the frequency of the note produced in the nth normal mode of vibration of the closed organ pipe is: $$ D_n = (2n+1)\frac{ν}{4L} = (2n+1)D₁ $$ - $$D_n = (2n+1)D₁ $$ ## Position of Nodes - x = 0, $\frac{L}{2n+1}$, $\frac{3L}{2n+1}$,$\frac{5L}{2n+1}$,........$\frac{(2n+1)L}{2n+1}$. - $$\frac {4L}{2n+1} =\frac{4L}{(2n+1)} = \frac{2L}{2n+1}$$ - $$\frac {4L}{2n+1} =\frac{4L}{(2n+1)} = \frac{2L}{1+2(n-1)} $$ ## Position of Antinodes - x = $\frac{L}{2n+1}$, $\frac{3L}{2n+1}$, $\frac{5L}{2n+1}$, ........$\frac{(2n+1)L}{2n+1}$. - The antinode lies between two nodes: - $$\frac{0+L}{ 2}= \frac{L}{2} $$ - $$\frac{2L + 4L}{2 (2n+1)}= \frac{6L}{ 2(2n+1)} = \frac{3L}{2n+1} $$ ## Standing Waves in Open Organ Pipes: Analytical Treatment - An Open organ pipe is open at both ends. - Therefore, an antinode is formed at each end. - $$ λ = \frac{2L}{n