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Velocity and acceleration of the medium

As seen in [link] , the wave speed is constant and represents the speed of the wave as it propagates through the medium, not the speed of the particles that make up the medium. The particles of the medium oscillate around an equilibrium position as the wave propagates through the medium. In the case of the transverse wave propagating in the x -direction, the particles oscillate up and down in the y -direction, perpendicular to the motion of the wave. The velocity of the particles of the medium is not constant, which means there is an acceleration. The velocity of the medium, which is perpendicular to the wave velocity in a transverse wave, can be found by taking the partial derivative of the position equation with respect to time. The partial derivative is found by taking the derivative of the function, treating all variables as constants, except for the variable in question. In the case of the partial derivative with respect to time t , the position x is treated as a constant. Although this may sound strange if you haven’t seen it before, the object of this exercise is to find the transverse velocity at a point, so in this sense, the x -position is not changing. We have

y ( x , t ) = A sin ( k x ω t + ϕ ) v y ( x , t ) = y ( x , t ) t = t ( A sin ( k x ω t + ϕ ) ) = A ω cos ( k x ω t + ϕ ) = v y max cos ( k x ω t + ϕ ) .

The magnitude of the maximum velocity of the medium is | v y max | = A ω . This may look familiar from the Oscillations and a mass on a spring.

We can find the acceleration of the medium by taking the partial derivative of the velocity equation with respect to time,

a y ( x , t ) = v y t = t ( A ω cos ( k x ω t + ϕ ) ) = A ω 2 sin ( k x ω t + ϕ ) = a y max sin ( k x ω t + ϕ ) .

The magnitude of the maximum acceleration is | a y max | = A ω 2 . The particles of the medium, or the mass elements, oscillate in simple harmonic motion for a mechanical wave.

The linear wave equation

We have just determined the velocity of the medium at a position x by taking the partial derivative, with respect to time, of the position y . For a transverse wave, this velocity is perpendicular to the direction of propagation of the wave. We found the acceleration by taking the partial derivative, with respect to time, of the velocity, which is the second time derivative of the position:

a y ( x , t ) = 2 y ( x . t ) t 2 = 2 t 2 ( A sin ( k x ω t + ϕ ) ) = A ω 2 sin ( k x ω t + ϕ ) .

Now consider the partial derivatives with respect to the other variable, the position x , holding the time constant. The first derivative is the slope of the wave at a point x at a time t ,

slope = y ( x , t ) x = x ( A sin ( k x ω t + ϕ ) ) = A k cos ( k x ω t + ϕ ) .

The second partial derivative expresses how the slope of the wave changes with respect to position—in other words, the curvature of the wave, where

curvature = 2 y ( x , t ) x 2 = 2 2 x ( A sin ( k x ω t + ϕ ) ) = A k 2 sin ( k x ω t + ϕ ) .

The ratio of the acceleration and the curvature leads to a very important relationship in physics known as the linear wave equation    . Taking the ratio and using the equation v = ω / k yields the linear wave equation (also known simply as the wave equation or the equation of a vibrating string),

2 y ( x , t ) t 2 2 y ( x , t ) x 2 = A ω 2 sin ( k x ω t + ϕ ) A k 2 sin ( k x ω t + ϕ ) = ω 2 k 2 = v 2 ,

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
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John Reply
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Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
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David Reply
what is viscosity?
David
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emma Reply
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Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
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Adjanou
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Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
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Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
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Reofrir Reply
Practice Key Terms 4

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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