Homework Help: Mechanical Waves

1. May 3, 2005

sghaussi

Hello again! I have a few questions regarding mathematical descriptions of waves.

what does it mean when a wave is propagating to the left? and how would i go about drawing that? do i just put a negative sine in front of the function? (for example sin(x) is original, -sin(x) is propagating to the left?

Another question ...so I have the equation: y(x,t) = Acos[2pi/lamda(x-vt)] which is the wave function equation.

how would i go about solving the expression for the transverse velocity and maximum speed (of particle on string for example).

I don't want answers, just help starting this problem. I'm not a physics major or anything so if you dont mind using simpler explanations (though i appreciate any help you can give me!!)

and lastly...

A continuous succession of sinusoidal wave pulses are produced at one end of a very long string and travels along the length of the string. I am given the frequency, amplitude, and wavelength.

I'm trying to work with the function: y(x,t) = Acos2pi[(x/lamda) - (t/T)]

I know: T = 1/f & lamda = v/f & w=2f*pi=vk

I plug in all of my data for A, x, lamda, and T. However, I want to find the amount of time needed for the wave to travel a certain distance (which is also given) how would I go about doing that? Do i sent two functions equal to each and solve for t? I tried, but it doesn't make sense to me and I don't even know if that is a legal move.

Thanks you guys!

2. May 3, 2005

nishant

for the first question,if the wave has the same phase or differs by 2npi,and is travelling towards left then replace x by -x,if it has a phase change of pi,then replace by x and the whole equation by {-}

3. May 3, 2005

nishant

to calculate the speed of wave just differentiate y w.r.t t.

4. May 3, 2005

sghaussi

so to find the velocity, just find the derivative of y? but with respect to what? And what does y w.r.t t. stand for? (okay i think i remember this method from calculus. so the main equation is just the position, then derivative of that would be velocity, and derivative of that would be acceleration right? am i thinking of the correct thing?)

5. May 3, 2005

nishant

ya you r thinking the right thing,derivative means the rate of change of a thing wrt to the other thing

6. May 4, 2005

sghaussi

okay then this is my position function (for a wave):

y(x,t) = A cos [ (2pi/lamda) * (x-vt) ]

I want to find the derivative for this because I'm trying to find an expression for the transverse velocity. I want to differentiate with respect to t only? so does A, 2pi/lamda, and v turn into constants? does x turn into a constant as well?\

when i try and find the derivative, should I apply the chain rule as well as the product rule?

Last edited: May 4, 2005
7. May 4, 2005

sghaussi

okay, I found the derivative with respect to "t". and this is what i have:

10*(pi/lamda)*v*sin[ (2pi/lamda*(1-vt)) ] does this look reasonable?

if this is the correct function for velocity, how would i go about finding the maximum speed?

Last edited: May 4, 2005
8. May 4, 2005

OlderDan

I think it would be best to rewind this a bit, so I will start at the top.

A wave propegates in the direction that x must change to keep the value of y constant. Look at your wave equation.

y(x,t) = Acos[2pi/lamda(x-vt)]

Look at a point where y has its maximum possible value. That is where cos[2pi/lamda(x-vt)]=1 and so y = A. We call such a point a crest of the wave. One such place is where x-vt = 0. Other points that are crests are where x - vt = n*lamda. These points are separated by lamda, the wavelength of the wave.

Time marches forward and there is no stopping it. What must x do to keep x-vt = 0 (actually, any constant value, but let's use zero)? It must be x = vt at all times, so the postion of the crest of the wave (point where y = A) is constantly changing and it is increasing at the rate v. v is the velocity at which the crests of the wave move.

This wave is moving to the right because x must be constantly increasing. A wave moving to the left would have a sign change in the cosine function

y(x,t) = Acos[2pi/lamda(x+vt)]

To keep y = A, x would have to always be equal to -vt. x = -vt means the wave is moving to the left with speed v.

This speed is not the speed of the particles of the string. It is the speed that the crests move. The particles on the string move transversely to the direction of the wave, and their velocity is found by taking the derivative of y(x,t) at some constant value of x.

dy/dt = -(2pi/lamda)Avsin(x+vt)

If we choose as one example point x = 0 this is

dy/dt = -(2pi/lamda)Avsin(vt)

Each little piece of the string oscillates like a harmonic oscillator, with a maximum transverse speed of (2pi/lamda)Av. The effect of different values of x is to change the phase of the motion at different points. Every position x along the string oscillates like a harminic oscillator, but adjacent pieces are slightly out of phase. If you took a snapshot of the string at one moment in time, it would be in the shape of a sine wave.

If you get these basics, i think you can go on to the later questions.

9. May 4, 2005

sghaussi

thank you Older Dan for the clear explanation.

I found the maximum speed to be A*omega from the equation:

A*omega*sin(kx-omega*t)

what does propogation speed v mean?

10. May 4, 2005

OlderDan

v is the velocity at which the crests of the wave move.

11. May 4, 2005

Hybird

For the last question you are asked to find the Time required for the wave to propogate (move) a certain distance. What has been explained is how to find the velocity of points, given a fixed position 'x', moving up and down perpendicular to wave propogation. And it is obvious that this velocity is not constant, it changes wrt (with respect to) time. Thats why you needed to differentiate.

What your looking for is the speed that the crests move paralell to the propagation of the wave. The nice thing about this is that all waves have constant velocity in the direction of propogation. And so you can use a quite simple formula here ie: D=VT. You know the distance, and you will know the velocity.

For a travelling wave it is quite easy to pick out these things. Take a look at the formula once more.

Y(x,t) = A sin [2pi/lambda(x-vt)]

This can also be written as

Y(x,t) = A sin (kx-wt)

where k = 2pi/lambda and w = 2pi/T

And this k is called the wave number.
(I'm assuming you messed up and put 'cos' instead of 'sin' in your equations?)

Now you know that 2pi/lam(x-vt) = kx-wt
so how did vt turn into wt? well it was simply multiplied by '2pi/lam' and thus:

wt=2pi/lambda*vt

cancel what you can, and solve for v. And you have all the information to solve the question.

EDIT: wait, I just actually looked at the question. That is really odd, the way it is setup, so I dont know if you could apply the wave number K the same way.. Sorry my bad

EDIT 2: Ok I looked it over. It is set up the same as the travelling wave, it just doesnt look like it. But I'm still iffy about that 'cos' thinking it should be a 'sin'. But heres what I came up with

we will neglect the A*cos for the time being. And so we have:

2pi(x/Lambda-t/T) bring the 2pi in = (2pi*x/lambda - 2pi*t/T)

and we know 2pi/lambda = k so

(kx-2pi*t/T)

and what does T=? well 1/f, so 1/T = f

(kx-2pi*f*t) and omega = 2pi*f

so we have : (kx-wt) standard for a travelling wave. And so yes you can use that first bit I applied. Just work through it.

Last edited: May 4, 2005
12. May 5, 2005

OlderDan

The use of cosine or sine to represent the wave is equivalent. The sine function and the cosine function are the same function with one or the other being shifted a phase distance $\pm\pi/2$ relative to the other. Whether you have $sin(kx - \omega t)$ or $cos(kx - \omega t)$ or $sin(kx - \omega t+\theta)$ or $cos(kx - \omega t+\theta)$ (where $\theta$ is a phase constant) depends on how you choose to describe the waveform at t = 0. The substitution of k and $\omega$ is correct, and it does make the represntation look cleaner.

Last edited: May 5, 2005
13. May 5, 2005

Hybird

Ya I thought about the whole 1/2 pi thing earlier today, god I spend to much time with physics on my head.