Is the wave phase of a sound wave always positive?

Question

I guess this is a pretty basic question, but I am not certain about its answer (since I am still studying in high school).

Let's say we are given the equation of the following sound wave and we are also asked to find the velocity of a particle whose x=1 m and t=8 s. The equation is given in SI units:

$$ y( x,t) =3\ sin\ \left( \ \frac{\pi }{4} \ t\ -\ 4\pi \ x\ \right) $$

If we try to find the velocity of the particle by taking the derivative of y(x,t) with respect to time, we get the following result:

$$ v( x,t) =y'(x,t)=\frac{3\pi }{4} \ cos\ \left( \ \frac{\pi }{4} \ t\ -\ 4\pi \ x\ \right) $$

Then, substituting the given values: $$ v( 1,8) =\frac{3\pi }{4} \ cos\ \left(-2\pi\right)=\frac{3\pi }{4}\ m/s $$ However, the wave velocity, which would be v = 0.0625 m/s, when multiplied by t = 8 s, gives us 0.5 m. Considering that the particle is x = 1 m away from the wave origin, the particle shouldn't have any particle velocity at all, should it? That would be because the wavefront hasn't reached the point where the particle is yet. Why is it that the two results don't match?

According to my book, the correct solution would be that the particle is not moving at all. But I wonder why my reasoning does not work.

I hope I've expressed my problem clearly. Thanks in advance.

Answer 1

You're equation does not model the start of the wave.

A sine function is defined for all input values from negative infinity to infinity. So your equation models the wave in a way that it has always been propagating and will always continue to propagate. You can even put in negative time values and get a value out of it.

You would have to adopt your equation with a step function. The definition of the step function is below. $$ u(t) = \begin{cases} 0~~~~~t < 0 \\ 1~~~~~t \ge 0\end{cases}$$

It is 0 for all negative values and 1 for all values greater or equal to 0.

The result would be: $$ y(x,t) = 3 \sin \left(\frac{\pi}{4}t - 4\pi x\right) \cdot u\left(t - \frac{x}{0.0625}\right)$$

The multiplication with the step function ensures, that your equation will result in $ y = 0 $ if the wave has not reached the point yet.

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Note that this equation still does not correctly catch the case if you put in a negative $x$ coordinate. This equation behaves like the wave started an ininite time ago at an infinite negative distance with the constraint that it reaches $x=0$ at $t = 0$.

For a wave that start in point $x = 0$ at time $t = 0$ you would have to use:

$$ y(x,t) = 3 \sin \left(\frac{\pi}{4}t - 4\pi x\right) \cdot u\left(t - \frac{x}{0.0625}\right)\cdot u\left(x\right)$$

This ensures the function is always 0 for all $ x < 0$

Answer 2

Sound is a pressure wave. An object vibrates back and forth. As it moves forward, it pushes air away creating a momentary region of high pressure. As it retreats, it creates a partial vacuum and air flows back. Air molecules near the object move back and forth. You can see this in the $sin(\frac{\pi}{4}t)$ part of the formula.

The air vibrating because of the object acts something like the object. It sets air a little farther away vibrating. The wave propagates. You can see this in the $sin(-4\pi x)$ part of the formula. The vibration continues for large x.

In the formula, the vibration lasts for all time and is happening at all x. There is no t so small or x so big that you don't find vibration. Obviously this doesn't describe all real sounds, where they start, continue for a while, and then stop. But it is a good place to start learning about waves.

The velocity equation you derived is like the x equation. Particles vibrate everywhere and at all times. They have a back and forth motion, where a positive $v$ means moving away, and a negative $v$ means moving back. Note that on the average, $v = 0$. A particle vibrates and keeps returning to the same positions over and over. It doesn't move like a wind blowing. The $x$ value is the displacement from the position it would stay at if there was no sound.

So $v(1,8)$ is the velocity of the particle that is at distance 1 and time 8. It gives the particle velocity at that time and place. There will be times when the velocity at that place is $0$, and times when it is not.

Answer 3

Without actually reading the question, it's hard to speculate as to why the book's solution doesn't agree. One possibility is that the book is actually giving you a pressure wave, instead of a displacement wave. The two are actually linked:

$$p(x,t) = C\frac{d}{d x}s(x,t)$$

where C is just a constant which we don't care about right now. In your context, if y(x,t) is a pressure wave, the displacement of your particle is:

$$s(x,t) = A\ cos\ (\frac{\pi}{4}t−4πx)$$

where A is the result of the integration. What matters is that now your displacement is a cosine wave instead of a sine wave. When you calculate the velocity of the particle:

$$s'(x,t) = -\frac{3\pi}{4}A\ sin\ (\frac{\pi}{4}t−4πx)$$

and evaluating this given your data returns 0. As someone else said, all of the waves which you define as sines or cosines have always existed and propagated in all of space (since they are defined for every x and t) so your argument is not really correct. This said, right now and without reading the question and knowing the context, I cannot think any other reason for the velocity being 0.