# Homework Help: Error in solutions or am I doing something wrong?

1. Feb 6, 2010

### Void123

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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).

2. Feb 6, 2010

### willem2

I think you have an extra A in that answer.

3. Feb 6, 2010

### jdwood983

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

$$\frac{dz}{dt}=A\frac{d}{dt}\cos[\omega t]=-A\omega\sin[\omega t]$$

$$\frac{d^2z}{dt^2}=-A\omega\frac{d}{dt}\sin[\omega t]=-A\omega^2\cos[\omega t]=-\omega^2 z$$

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

4. Feb 6, 2010

### 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..

5. Feb 6, 2010

### jdwood983

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

6. Feb 6, 2010

### Void123

The argument is $$cos(kx -\omega t)$$

7. Feb 6, 2010

### jdwood983

Hmm, still doesn't seem to change anything

$$\frac{dz}{dt}=A\frac{d}{dt}\cos[kx-\omega t]=\omega A\sin[kx-\omega t]$$

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

$$\frac{d^2z}{dt^2}=\omega A\frac{d}{dt}\sin[kx-\omega t]=-\omega^2A\cos[kx-\omega t]=-\omega^2z$$

8. Feb 6, 2010

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

9. Feb 6, 2010

### 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?

10. Feb 6, 2010

### 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 $$-\omega^{2} z(x, t)$$

11. Feb 6, 2010

### jdwood983

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

12. Feb 6, 2010

### Void123

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