Error in solutions or am I doing something wrong?

[Void123]

Homework Statement

This is quite simple, but for some reason I'm getting the wrong answer. I am given a longitudinal wave where the vertical displacement is given by z(x, t) = Acos(...). I must calculate the vertical acceleration at some time t and position x.

Homework Equations

The Attempt at a Solution

I just differentiate twice with respect to time and get [-\omega^{2}A][z(x,t))]. Correct?

This is not the first time I have encountered a solution error (if it is the case).

 

[willem2]

I think you have an extra A in that answer.

 

[jdwood983]

Assuming the (...) argument of the cosine is just \omega t, then

<br /> \frac{dz}{dt}=A\frac{d}{dt}\cos[\omega t]=-A\omega\sin[\omega t]<br />

<br /> \frac{d^2z}{dt^2}=-A\omega\frac{d}{dt}\sin[\omega t]=-A\omega^2\cos[\omega t]=-\omega^2 z<br />

So willem is correct, you multiplied by an extra A.

 

[Void123]

Ah, yes sorry I did not mean to put that extra A there.

But it should read -\omega^{2}Acos(...)

This was my original answer and its still wrong.

Ehhh..

 

[jdwood983]

What is the argument of cosine? You keep putting "..." in there, just write it out for me to help you.

 

[Void123]

The argument is cos(kx -\omega t)

 

[jdwood983]

Hmm, still doesn't seem to change anything

<br /> \frac{dz}{dt}=A\frac{d}{dt}\cos[kx-\omega t]=\omega A\sin[kx-\omega t]<br />

because you have a -\omega coming from the argument and then a negative sign coming from the derivative of cosine.

<br /> \frac{d^2z}{dt^2}=\omega A\frac{d}{dt}\sin[kx-\omega t]=-\omega^2A\cos[kx-\omega t]=-\omega^2z<br />

What is telling you your answer is wrong, your textbook, website, intuition, etc??

 

[Void123]

It is a website. I considered the possibility of my syntax being wrong, but I went over it and couldn't find any error. It tells us to express the answer in terms of specific variables, which I did.

I don't know. This isn't the first time the website has made a mistake, so I wouldn't be surprised if its happened again.

 

[jdwood983]

Did you write it as -\omega^2z or -\omega^2A\cos[kx-\omega t]? If you wrote the latter, I haven't a clue as to what the website would be asking you, unless you were given some constants to use?

 

[Void123]

Copied and pasted from the site:

"Express the vertical acceleration in terms of \omega, g, k, A, and the independent variables x and t."

Keep in mind that \omega = \sqrt{g k}

I came up with a_{z} (x, t) = - g k Acos(kx - \omega t) (as my officially submitted answer)

which is equal to -\omega^{2} z(x, t)

 

[jdwood983]

That looks like the correct formula to me, I do not know why you would be getting it wrong.

 

[Void123]

Thanks. All I needed was a secondary supporting opinion. I will take it up with the instructor then.