Particle displacement by a speaker in a gas tube

[beyondlight]

A speaker is generating waves in a gas that is closed in a tube. The tubes closed end is at x=0 and the speaker is at x=1. The motion of the membrane is given by

$$s=s_0 \cdot sin(\omega t)$$

How does the particle displacement vary with x and t? The wavelength is $$\lambda$$I do not know how to solve it. But i thougt of expressing it like this:

$$s_1=s_0 sin(kx + \omega t)$$ (wave due to speaker oscillation, traveling from x=1 to x=0)

$$s_2=s_0sin(kx - \omega t)$$ (reflected wave at x=0 traveling to x=1)

The problem is that the reflected wave is reflected again and again...

 

[fizika_kz]

there is standing wave. do you know what it is?

 

[beyondlight]

Yes i know what it is. If we solve it as a standing wave problem then we have:

$$s_{tot}=2s_0cos(\omega t)sin(kx)$$

Boundary condition: $$s_{tot}=0$$ at x=0


According to the answer in my book it is a bit different, but x and t are separate from each other in the sinusodial functions as my suggestion above.

Any hint for me?

please

 

[Philip Wood]

You've got the right idea, but you need to be a bit more flexible with your choice of phases of progressive waves. You have to choose them so that their sum conforms to the boundary conditions at x=0 and x=1.

Yours fit at x=0, because sin(kx) =0, and s₁ + s₂ must indeed be zero next to the wall.

But things aren't right at x=1. The speaker diaphragm moves according to sin(wt), whereas your s₁ + s₂ varies as cos(wt). May I suggest you try the progressive waves s₁ = Acos(kx+wt) and s₂ = -Acos(kx-wt). Then express the difference of two cosines as a product. It comes out as a product of sines, so you can meet both boundary conditions.

 

[beyondlight]

I don't really know how to do...

If we put a boundary condition at x=1 as:

s=2scos(wt)sin(k)

now what?

But i recognized that this should also be a fact:

2scos(wt)sin(k)=sin(wt) this should also help i think

 

[Philip Wood]

Did you add the two progressive waves that I recommended you try? Use the standard trig formula cos($\vartheta$) - cos($\varphi$) = 2sin[($\vartheta$ +$\varphi$)/2] sin[($\vartheta$ -$\varphi$)/2].

You should then find that s₁ +s₂ is the product of an x-dependent sine and a t-dependent sine. s₁ +s₂ automatically disappears when x = 0, just as it should at a wall. It also agrees with the motion of the membrane at x = 1, as given in the question, provided that A =...