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While I was studying beats, I tried to find a displacement function of any particle in the most generalized form. I ended up with $$y=2A\sin(\pi(t-x/v)(f_1+f_2))\cos(\pi(t-x/v)(f_1-f_2)).$$

Now, given such a function wherein you cannot compare to a general equation, how do you find the wave speed and the phase speed? And, I think, some clarification on how "group velocity" applies to such a function or phenomenon would be appreciated .

PS: Can we first break down this function using Fourier analysis and then maybe simplify things a bit?

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For the example function you state, group and phase speed are identical. You are beating two waves that have same speed. Try beating two waves with different speed.

OK, say we are adding two waves, $$y=A\sin(2\pi f_1(t-x/v_1))+A\sin(2\pi f_2(t-x/v_2))$$

Note that both waves have different frequencies and velocities. Then use $$\sin(A)+\sin(B) = 2\sin((A+B)/2)\cos((A-B)/2)$$

Do some algebra and you will get $$y = 2A\sin(\pi (f_1+f_2)(t-x/v_+))\cos(\pi(f_1-f_2)(t-x/v_-))$$ where $$v_+ = v_1 v_2(f_1 + f_2) / (v_2 f_1 + v_1 f_2)$$ $$v_- = v_1 v_2(f_1 - f_2) / (v_2 f_1 - v_1 f_2)$$

Now if both waves have the same speed, $ v_1 = v_2 = v$, then you get the formula you started with, with both the high frequency and low frequency components moving at the same speed. If $ v_1 \neq v_2$ then $v_+$ would be considered the "phase velocity" and $v_-$ the "group velocity".

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  • $\begingroup$ By 'speed' do you mean frequency? $\endgroup$
    – Kitchi
    Apr 10, 2013 at 19:49
  • $\begingroup$ No, by 'speed' I mean speed. Both the high frequency component an low frequency envelope are translated at the same speed 'v'. $\endgroup$
    – Horatio
    Apr 10, 2013 at 20:10
  • $\begingroup$ well then i don't have an example of that, it would be nice if you could give one. $\endgroup$ Apr 11, 2013 at 2:04

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