# Amplitude in an equation

1. Feb 6, 2004

### BLUE_CHIP

What does the $$A$$ stand for in the equation:

$$y=A\sin{(kx-t\omega)}$$

CHEERS :)

2. Feb 6, 2004

### chroot

Staff Emeritus
Amplitude.

- Warren

3. Feb 6, 2004

### himanshu121

max value of the displacement from the mean position

4. Feb 6, 2004

### BLUE_CHIP

thanks :)

whats the relationship between $$k$$ and the wavelength of the wave

5. Feb 6, 2004

### himanshu121

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

6. Feb 6, 2004

### chroot

Staff Emeritus
Think about it. If x is the displacement along a taught string, the wavelength of a wave on that string is the distance between successive crests or troughs.

All sine waves repeat every 2 pi radians.

When $x = \lambda$, you want the argument to be $2 \pi$.

Try rewriting the first term (the term with the x) as:

$$\frac{2 \pi x}{\lambda}$$

You'll see that when $x = \lambda$, the entire expression is $2 \pi$ -- exactly one period. This is the right expression.

Therefore, if you want to simplify that expression by bringing in a new symbol k, k must be

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

- Warren

Last edited: Feb 6, 2004
7. Feb 6, 2004

### BLUE_CHIP

Score! thanks Boudoir

8. Feb 6, 2004

### BLUE_CHIP

Bummer hit a brick wall again. check this out:

for the equation $$y=A\sin{(kx-t\omega)}$$ find a relationship between $$\omega$$ and the time period $$T$$ of the wave.

when $$t=T$$ $$y=0$$ and $$x=0$$

therefore:

$$A\sin{(-T\omega)}=0$$

but then what?

9. Feb 6, 2004

### chroot

Staff Emeritus
Don't you have a textbook?

$$\omega = 2 \pi f$$

$$T = \frac{1}{f}$$

$$T = \frac{2 \pi}{\omega}$$

- Warren

10. Feb 6, 2004

### GRQC

I think you're doing his homework for him.

11. Feb 7, 2004

### himanshu121

I dont find it as Homework.

Anyway He is reaching the conclusions and thats the bottom line

12. Feb 7, 2004

### BLUE_CHIP

Thanks saved my life.