# Oscillation of soundwave

1. May 21, 2015

### nuuskur

Given a soundwave with wavelength $\lambda = 0,628 m$ and a period $T = 2 ms$.
The stopper is started at the exact moment when the wave is at its minimum, call it $-A$. After $3.5ms$ and $0,157m$ from the point of origin, the wave has reached its maximum, $A$.

Why is it so?
According to my reasoning the wave has reached 0 at that time. It takes the $2ms$ to reach minimum again and then with the remaining $1.5ms$ it reaches 0 with the first $0.5ms$, then the maximum after $1ms$ and then 0 again after $1.5ms$, but this is incorrect.

Edit: I would understand if we were given just the distance it has traveled in which case it would have travelled exactly a fourth from the point of origin and therefore reaching its maximum, but then why the $3.5ms$?

Last edited: May 21, 2015
2. May 21, 2015

### sunmaggot

think of the appearance of the wave, say it moves towards you, particles are propagating along y direction. When you are at 0.157m away, at t=0, you should see this point is y=0. After 0.5ms, the particle will have y= -A, then after 1ms, the particle will have y=A. Then after one more period, 2ms, that is totally 3.5ms, that particle will have y=A, which is maximum.
so it seems that it has no problem.

Edit: sound wave is not transfer wave, but I used transfer wave to explain, but they should have same result.

3. May 21, 2015

### nuuskur

Doesn't it say at t=0 y=-A?

4. May 21, 2015

### sunmaggot

that one is for particle position at stopper, but remember that 0.157m? the position 0.157m away from stopper has particle at y=0 when t=0

5. May 21, 2015

### nasu

You can just use the formula for the phase, no need to guess.
You have $$\phi =\omega t - k x =2 \pi (\frac{t}{T}-\frac{x}{\lambda})$$.
For t=0 and x=0 you have $$\phi_1 =0$$
For the values given in OP, you have $$\phi_2 = 2 \pi (\frac{3.5}{2}-\frac{0.175}{0.628})=3 \pi$$
Multiples of 2Π do not change the wave at all. So this is an actual change of Π which means that the two events are in opposition of phase. So from -A it goes to +A.

If you still want to describe it "in two pieces", the x2=0.157m is one quarter wavelength. So at t=0, and x1=0, the wave at x2=0.175 m goes through zero, towards negative values.
First time it will reach +A will be after 3/4 of a period. Next time it will reach the same value will be after 3/4+1 period or 1.75 periods or 3.5 ms.

6. May 22, 2015

### theodoros.mihos

Like @nasu, your wave equation is:
$$y(x,t) = A\sin(1000\pi{t}-10x)$$

Edit: if demand to calculate values for large time (t>10T), some programs makes truncation error mistakes.