# 16.2 Mathematics of waves  (Page 2/11)

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To construct our model of the wave using a periodic function, consider the ratio of the angle and the position,

$\begin{array}{ccc}\hfill \frac{\theta }{x}& =\hfill & \frac{2\pi }{\lambda },\hfill \\ \hfill \theta & =\hfill & \frac{2\pi }{\lambda }x.\hfill \end{array}$

Using $\theta =\frac{2\pi }{\lambda }x$ and multiplying the sine function by the amplitude A , we can now model the y -position of the string as a function of the position x :

$y\left(x\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(\frac{2\pi }{\lambda }x\right).$

The wave on the string travels in the positive x -direction with a constant velocity v , and moves a distance vt in a time t . The wave function can now be defined by

$y\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(\frac{2\pi }{\lambda }\left(x-vt\right)\right).$

It is often convenient to rewrite this wave function in a more compact form. Multiplying through by the ratio $\frac{2\pi }{\lambda }$ leads to the equation

$y\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(\frac{2\pi }{\lambda }x-\frac{2\pi }{\lambda }vt\right).$

The value $\frac{2\pi }{\lambda }$ is defined as the wave number    . The symbol for the wave number is k and has units of inverse meters, ${\text{m}}^{-1}:$

$k\equiv \frac{2\pi }{\lambda }$

Recall from Oscillations that the angular frequency    is defined as $\omega \equiv \frac{2\pi }{T}.$ The second term of the wave function becomes

$\frac{2\pi }{\lambda }vt=\frac{2\pi }{\lambda }\left(\frac{\lambda }{T}\right)t=\frac{2\pi }{T}t=\omega t.$

The wave function for a simple harmonic wave on a string reduces to

$y\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx\mp \omega t\right),$

where A is the amplitude, $k=\frac{2\pi }{\lambda }$ is the wave number, $\omega =\frac{2\pi }{T}$ is the angular frequency, the minus sign is for waves moving in the positive x -direction, and the plus sign is for waves moving in the negative x -direction. The velocity of the wave is equal to

$v=\frac{\lambda }{T}=\frac{\lambda }{T}\left(\frac{2\pi }{2\pi }\right)=\frac{\omega }{k}.$

Think back to our discussion of a mass on a spring, when the position of the mass was modeled as $x\left(t\right)=A\phantom{\rule{0.2em}{0ex}}\text{cos}\left(\omega t+\varphi \right).$ The angle $\varphi$ is a phase shift, added to allow for the fact that the mass may have initial conditions other than $x=\text{+}A$ and $v=0.$ For similar reasons, the initial phase is added to the wave function. The wave function modeling a sinusoidal wave, allowing for an initial phase shift $\varphi ,$ is

$y\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx\mp \omega t+\varphi \right)$

The value

$\left(kx\mp \omega t+\varphi \right)$

is known as the phase of the wave , where $\varphi$ is the initial phase of the wave function. Whether the temporal term $\omega t$ is negative or positive depends on the direction of the wave. First consider the minus sign for a wave with an initial phase equal to zero $\left(\varphi =0\right).$ The phase of the wave would be $\left(kx-\omega t\right).$ Consider following a point on a wave, such as a crest. A crest will occur when $\text{sin}\phantom{\rule{0.2em}{0ex}}\left(kx-\omega t\right)=1.00$ , that is, when $kx-\omega t=n\pi +\frac{\pi }{2},$ for any integral value of n . For instance, one particular crest occurs at $kx-\omega t=\frac{\pi }{2}.$ As the wave moves, time increases and x must also increase to keep the phase equal to $\frac{\pi }{2}.$ Therefore, the minus sign is for a wave moving in the positive x -direction. Using the plus sign, $kx+\omega t=\frac{\pi }{2}.$ As time increases, x must decrease to keep the phase equal to $\frac{\pi }{2}.$ The plus sign is used for waves moving in the negative x -direction. In summary, $y\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)$ models a wave moving in the positive x -direction and $y\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx+\omega t+\varphi \right)$ models a wave moving in the negative x -direction.

[link] is known as a simple harmonic wave function. A wave function is any function such that $f\left(x,t\right)=f\left(x-vt\right).$ Later in this chapter, we will see that it is a solution to the linear wave equation. Note that $y\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{cos}\left(kx+\omega t+\varphi \text{′}\right)$ works equally well because it corresponds to a different phase shift $\varphi \text{′}=\varphi -\frac{\pi }{2}.$

## Problem-solving strategy: finding the characteristics of a sinusoidal wave

1. To find the amplitude, wavelength, period, and frequency of a sinusoidal wave, write down the wave function in the form $y\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right).$
2. The amplitude can be read straight from the equation and is equal to A .
3. The period of the wave can be derived from the angular frequency $\left(T=\frac{2\pi }{\omega }\right).$
4. The frequency can be found using $f=\frac{1}{T}.$
5. The wavelength can be found using the wave number $\left(\lambda =\frac{2\pi }{k}\right).$

A body receives impulses of 24Ns and 35Ns inclined 55 degree to each other. calculate the total impulse
what is matar
The uniform boom shown below weighs 500 N, and the object hanging from its right end weighs 400 N. The boom is supported by a light cable and by a hinge at the wall. Calculate the tension in the cable and the force on the hinge on the boom. Does the force on the hinge act along the boom?
A 11.0-m boom, AB , of a crane lifting a 3000-kg load is shown below. The center of mass of the boom is at its geometric center, and the mass of the boom is 800 kg. For the position shown, calculate tension T in the cable and the force at the axle A .
Jave
what is the S.I unit of coefficient of viscosity
Derived the formula of Newton's law of universal gravitation Fg=G(M1M2)/R2
a non-uniform boom of a crane 15m long, weighs 2800nts, with its center of gravity at 40% of its lenght from the hingr support. the boom is attached to a hinge at the lower end. rhe boom, which mAKES A 60% ANGLE WITH THE HORIZONTAL IS SUPPORTED BY A HORIZONTAL GUY WIRE AT ITS UPPER END. IF A LOAD OF 5000Nts is hung at the upper end of the boom, find the tension in the guywire and the components of the reaction at the hinge.
what is the centripetal force
Of?
John
centripetal force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
centripetal force is the force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
I do believe the formula for centripetal force is F=MA or F=m(v^2/r)
John
I mean the formula is Fc= Mass multiplied by square of velocity all over the Radius of the circle
Sampson
Yes
John
The force is equal to the mass times the velocity squared divided by the radius
John
That's the current chapter I'm on in my engineering physics class
John
Centripetal force is a force of attraction which keeps an object round the orbit towards the center of a circle. Mathematically Fc=mv²/r
In Example, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m bel
A steel lift column in a service station is 4 meter long and .2 meter in diameter. Young's modulus for steel is 20 X 1010N/m2.  By how much does the column shrink when a 5000- kg truck is on it?
hi
Abdulrahman
mola mass
Abdulrahman
what exactly is a transverse wave
does newton's first law mean that we don't need gravity to be attracted
no, it just means that a brick isn't gonna move unless something makes it move. if in the air, moves down because of gravity. if on floor, doesn't move unless something has it move, like a hand pushing the brick. first law is that an object will stay at rest or motion unless another force acts upon
Grant
yeah but once gravity has already been exerted .. i am saying that it need not be constantly exerted now according to newtons first law
Dharmee
gravity is constantly being exerted. gravity is the force of attractiveness between two objects. you and another person exert a force on each other but the reason you two don't come together is because earth's effect on both of you is much greater
Grant
maybe the reason we dont come together is our inertia only and not gravity
Dharmee
this is the definition of inertia: a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.
Grant
the earth has a much higher affect on us force wise that me and you together on each other, that's why we don't attract, relatively speaking of course
Grant
quite clear explanation but i just want my mind to be open to any theory at all .. its possible that maybe gravity does not exist at all or even the opposite can be true .. i dont want a fixed state of mind thats all
Dharmee
why wouldn't gravity exist? gravity is just the attractive force between two objects, at least to my understanding.
Grant
earth moves in a circular motion so yes it does need a constant force for a circular motion but incase of objects on earth i feel maybe there is no force of attraction towards the centre and its our inertia forcing us to stay at a point as once gravity had acted on the object
Dharmee
why should it exist .. i mean its all an assumption and the evidences are empirical
Dharmee
We have equations to prove it and lies of evidence to support. we orbit because we have a velocity and the sun is pulling us. Gravity is a law, we know it exists.
Grant
yeah sure there are equations but they are based on observations and assumptions
Dharmee
g is obtained by a simple pendulum experiment ...
Dharmee
gravity is tested by dropping a rock...
Grant
and also there were so many newtonian laws proved wrong by einstein . jus saying that its a law doesnt mean it cant be wrong
Dharmee
pendulum is good for showing energy transfer, here is an article on the detection of gravitational waves: ***ligo.org/detections.php
Grant
yeah but g is calculated by pendulum oscillations ..
Dharmee
thats what .. einstein s fabric model explains that force of attraction by sun on earth but i am talking about force of attraction by earth on objects on earth
Dharmee
no... this is how gravity is calculated:F = G*((m sub 1*m sub 2)/r^2)
Grant
gravitational constant is obtained EXPERIMENTALLY
Dharmee
the G part
Dharmee
Calculate the time of one oscillation or the period (T) by dividing the total time by the number of oscillations you counted. Use your calculated (T) along with the exact length of the pendulum (L) in the above formula to find "g." This is your measured value for "g."
Dharmee
G is the universal gravitational constant. F is the gravity
Grant
search up the gravity equation
Grant
yeahh G is obtained experimentally
Dharmee
sure yes
Grant
thats what .. after all its EXPERIMENTALLY calculated so its empirical
Dharmee
yes... so where do we disagree?
Grant
its empirical whixh means it can be proved wrong
Dharmee
so cant just say why wouldnt gravity exists
Dharmee
the constant, sure but extremely unlikely it is wrong. gravity however exists, there are equations and loads of support surrounding the concept. unfortunately I don't have a high enough background in physics but have this discussion with a physicist
Grant
can u suggest a platform where i can?
Dharmee
stack overflow
Grant
stack exchange, physics section***
Grant
its an app?
Dharmee
there is! it is also a website as well
Grant
okayy
Dharmee
nice talking to you
Dharmee
***physics.stackexchange.com/
Grant
likewise :)
Grant
Gravity surely exist
muhammed
hi guys
Diwash
hi
muhammed
what is mathematics
What is the percentage by massof oxygen in Al2(so4)3
molecular mass of Al2(SO4)3 = (27×2)+3{(32×1)+(16×4)} =54+3(32+64) =54+3×96 =54+288 =342 g/mol molecular mass of Oxygen=12×16 =192 g/mol % of Oxygen= (molecular mass of Oxygen/ molecular mass of the compound)×100% =(192/342)×100% =19200/342% =56.14%
A spring with 50g mass suspended from it,has its length extended by 7.8cm 1.1 determine the spring constant? 1.2 it is observed that the length of the spring decreases by 4.7cm,from its original length, when a toy is place on top of it. what is the mass of the toy?
solution mass = 50g= 0.05kg force= 50 x 10= 500N extension= 7.8cm = 0.078m using the formula Force= Ke K = force/extension 500/.078 = 6410.25N/m
Sampson
1.2 Decrease in length= -4.7cm =-0.047m mass=? acceleration due to gravity= 10 force = K x e force= mass x acceleration m x a = K x e mass = K x e/acceleration = 6410.25 x 0.047/10 = 30.13kg
Sampson
1.1 6.28Nm-¹
Anita
1.2 0.03kg or 30g
Anita
I used g=9.8ms-²
Anita
you should explain how yoy got the answer Anita
Grant
ok
Anita
with the fomular F=mg I got the value for force because now the force acting on the spring is the weight of the object and also you have to convert from grams to kilograms and cm to meter
Anita
so the spring constant K=F/e where F is force and e is extension
Anita
mass=50g=50/1000 kg m=0.05kg extension=7.8 cm=7.8/100 e=0.078 m g=9.8 m/s² 1.1 F=ke k=F/e k=mg/e k=0.05×9.8/0.078 k=0.49/0.078 k=6.28 N/m 1.2 F=6.28e mg=6.28e m=6.28e/g e=4.7 cm =4.7/100 e=0.047 m=6.28×0.047/9.8 m=0.29516/9.8 m=0.0301 kg
In this first example why didn't we use P=P° + ¶hg where ¶ is density
Density = force applied x area p=fA =p = mga, then a=h therefore substitute =p =mgh
Hlehle
Hlehle
sorry I had a little typo in my question
Anita
Density = m/v (mass/volume) simple as that
Augustine
Hlehle vilakazi how density is equal to force * area and you also wrote p= mgh which is machenical potential energy ? how ?
Manorama
what is wave
who can state the third equation of motion
Alfred
wave is a distrubance that travelled in medium from one point to another with carry energy .
Manorama
wave is a periodic disturbance that carries energy from one medium to another..
Augustine
what exactly is a transverse wave then?
Dharmee