# 16.2 Mathematics of waves  (Page 4/11)

 Page 4 / 11

## 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

$\begin{array}{ccc}\hfill y\left(x,t\right)& =\hfill & A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)\hfill \\ \hfill {v}_{y}\left(x,t\right)& =\hfill & \frac{\partial y\left(x,t\right)}{\partial t}=\frac{\partial }{\partial t}\left(A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)\right)\hfill \\ & =\hfill & \text{−}A\omega \phantom{\rule{0.2em}{0ex}}\text{cos}\left(kx-\omega t+\varphi \right)\hfill \\ & =\hfill & \text{−}{v}_{y\phantom{\rule{0.2em}{0ex}}\text{max}}\phantom{\rule{0.2em}{0ex}}\text{cos}\left(kx-\omega t+\varphi \right).\hfill \end{array}$

The magnitude of the maximum velocity of the medium is $|{v}_{{y}_{\text{max}}}|=A\omega$ . 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,

$\begin{array}{cc}\hfill {a}_{y}\left(x,t\right)& =\frac{\partial {v}_{y}}{\partial t}=\frac{\partial }{\partial t}\left(\text{−}A\omega \phantom{\rule{0.2em}{0ex}}\text{cos}\left(kx-\omega t+\varphi \right)\right)\hfill \\ & =\text{−}A{\omega }^{2}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)\hfill \\ & =\text{−}{a}_{y\phantom{\rule{0.2em}{0ex}}\text{max}}\text{sin}\left(kx-\omega t+\varphi \right).\hfill \end{array}$

The magnitude of the maximum acceleration is $|{a}_{{y}_{\text{max}}}|=A{\omega }^{2}.$ 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:

${a}_{y}\left(x,t\right)=\frac{{\partial }^{2}y\left(x.t\right)}{\partial {t}^{2}}=\frac{{\partial }^{2}}{\partial {t}^{2}}\left(A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)\right)=\text{−}A{\omega }^{2}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right).$

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 ,

$\text{slope}=\frac{\partial y\left(x,t\right)}{\partial x}=\frac{\partial }{\partial x}\left(A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)\right)=Ak\phantom{\rule{0.2em}{0ex}}\text{cos}\left(kx-\omega t+\varphi \right).$

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

$\text{curvature}=\frac{{\partial }^{2}y\left(x,t\right)}{\partial {x}^{2}}=\frac{{\partial }^{2}}{{\partial }^{2}x}\left(A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)\right)=\text{−}A{k}^{2}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right).$

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 $v=\omega \text{/}k$ yields the linear wave equation (also known simply as the wave equation or the equation of a vibrating string),

$\begin{array}{cc}\hfill \frac{\frac{{\partial }^{2}y\left(x,t\right)}{\partial {t}^{2}}}{\frac{{\partial }^{2}y\left(x,t\right)}{\partial {x}^{2}}}& =\frac{\text{−}A{\omega }^{2}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)}{\text{−}A{k}^{2}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)}\hfill \\ & =\frac{{\omega }^{2}}{{k}^{2}}={v}^{2},\hfill \end{array}$

#### Questions & Answers

when I click on the links in the topics noting shows. what should I do.
can we regard torque as a force?
Torque is only referred a force to rotate objects.
SHREESH
thanks
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I need lessons on Simple harmonic motion
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Given two vectors, vector C which is 3 units, and vector D which is 5 units. If the two vectors form an angle of 45o, determine C D and direction.
AFLAX
ty
Sharath
CD=5.83 n direction is NE
Ark
state Hooke's law of elasticity
Hooke's law states that the extension produced is directly proportional to the applied force provided that the elastic limit is not exceeded. F=ke;
Shaibu
thanks
Aarti
You are welcome
Shaibu
thnx
Junaid
what is drag force
Junaid
A backward acting force that tends to resist thrust
Ian
solve:A person who weighs 720N in air is lowered in to tank of water to about chin level .He sits in a harness of negligible mass suspended from a scale that reads his apparent weight .He then dumps himself under water submerging his body .If his weight while submerged is 34.3N. find his density
Ian
The weight inside the tank is lesser due to the buoyancy force by the water displaced. Weight of water displaced = His weight outside - his weight inside tank = 720 - 34.3 = 685.7N Now, the density of water = 997kg/m³ (this is a known value) Volume of water displaced = Mass/Density (next com)
Sharath
density or relative density
Shaibu
density
Ian
Upthrust =720-34.3=685.7N mass of water displayed = 685.7/g vol of water displayed = 685.7/g/997 hence, density of man = 720/g / (685.7/g/997) =1046.6 kg/m3
1046.8
R.d=weight in air/upthrust in water =720/34.3=20.99 R.d=density of substance/density of water 20.99=x/1 x=20.99g/cm^3
Shaibu
Kg /cubic meters
Shaibu
Upthrust = 720-34.3=685.7N vol of water = 685.7/g/density of water = 685.7/g/997 so density of man = 720/g /(685.7/g/997) =1046.8 kg/m3
is there anyway i can see your calculations
Ian
Upthrust =720-34.3=685.7
Upthrust 720-34.3
=685.7N
Vol of water = 685.7/g/997
Hence density of man = 720/g / (685.7/g/997)
=1046.8 kg/m3
so the density of water is 997
Shaibu
Yes
Okay, thanks
Shaibu
try finding the volume then
Ian
Vol of man = vol of water displayed
I've done that; I got 0.0687m^3
Shaibu
okay i got it thanks
Ian
u welcome
Shaibu
HELLO kindly assist me on this...(MATHS) show that the function f(x)=[0 for xor=0]is continuous from the right of x->0 but not from the left of x->0
I do not get the question can you make it clearer
Ark
Same here, the function looks very ambiguous. please restate the question properly.
Sharath
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Take a paper. put a point (name is A), now draw a line in the South east direction from A. Assume the line is 25 km long. that is the first stop (name the second point B) From B, turn 60 degrees to the north of East and draw another line, name that C. that line is 40 km long. (contd.)
Sharath
Now, you know how to calculate displacements, I hope? the displacement between two points is the shortest distance between the two points. go ahead and do the calculations necessary. Good luck!
Sharath
thank you so much Sharath Kumar
Liteboho
thank you, have also learned alot
Duncan
No issues at all. I love the subject and teaching it is fun. Cheers!
Sharath
cheers!
Liteboho
cheers too
Duncan
hii
Lakshya
hii too
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haye
Siciid
yes
Siciid
yes
Lakshya
shggggg
Lakshya
you mean
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solution problems
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yes
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0
Keshav
what is soln..
Keshav
o
Duncan
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Sharath
Some corrections to my old post. Time taken to reach ground = 2*v*Sin (∆)/g R = (2*v²*Sin(∆)*Cos(∆))/g I put the g in the numerator by mistake in my old post. apologies for that. R on moon = (R on Earth)/(0.166)
Sharath
state Newton's first law of motion
Every body will continue in it's state of rest or of uniform motion in a straight line, unless it is compelled to change that state by an external force.
Kumaga
if you want this to become intuitive to you then you should state it
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changing the state of rest or uniform motion of a body
koffi
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what is a vector
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103.077640640442km/h
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state Newton's first law of motion
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Alem
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Govindsingh By    By Abishek Devaraj     