What is the resultant frequency if two similar waves superpose?

[Bruce_Pipi121]

Homework Statement

Consider two sound waves travelling with the same speed and amplitude but having similar but slightly different wavelengths, $λ_1$ and $λ_2$, and angular frequencies, $ω_1$ and $ω_2$. The two waves are described with the functions $$y_1 (x,t) = A cos (\frac{2\pi x}{\lambda_1}- \omega_1 t)$$ $$y_2 (x,t) = A cos (\frac{2\pi x}{\lambda_2}- \omega_2 t)$$
a) What is the speed v in terms of the angular frequencies and wavelengths?
b) Sketch y1 + y2 as a function of x, at some time t.
c) If you stood in the path of these sound waves, what frequency would you hear (assuming you can hear it)?
d) What is the distance between points where the sound disappears?

Relevant Equations

$$cos x +cos y = 2 cos{\frac{x+y}{2}}cos{\frac{x-y}{2}}$$

for a) I have $v = \lambda f= \frac{\omega \lambda}{2 \pi}$
for c) and d) I denote $\frac{2\pi}{\lambda_1} = k_1$ and $\frac{2\pi}{\lambda_2} = k_2$ assuming $k_1 > k_2$.
so using the triangular identity I got $$y_1 + y_2 = 2A cos(\frac{k_1+k_2}{2} x - \frac{\omega_1 + \omega_2}{2} t) cos(\frac{k_1-k_2}{2} x - \frac{\omega_1 - \omega_2}{2} t)$$
To simplify the equation I denote $K_1 = \frac{k_1+k_2}{2}$, $K_2 = \frac{k_1-k_2}{2}$, $\Omega_1 = \frac{\omega_1+\omega_2}{2}$ and $\Omega_2 = \frac{\omega_1 - \omega_2}{2}$, so $$y_1 + y_2= 2A cos(K_1 x - \Omega_1 t)cos(K_2 x - \Omega_2 t)$$ and since $k_1 \approx k_2$ so $K_1 >> K_2$ and $\omega_1 \approx \omega_2$ so $\Omega_1 >> \Omega_2$
now from the graph I can solve for d) distance is $\frac{\pi}{K_2}$

but I am not sure about c) is the frequency we would hear is $f = \frac{\Omega_1}{2 \pi}$, why?

 

[anuttarasammyak]

In your notebook photo attached, replace x with t and K with $\Omega$. You see or hear sound of frequency $\Omega_1$ with amplitude or loudness changing with frequency $\Omega_2$.

 

[Bruce_Pipi121]

In your notebook photo attached, replace x with t and K with $\Omega$. You see or hear sound of frequency $\Omega_1$ with amplitude or loudness changing with frequency $\Omega_2$.

Thank you. I think it makes sense. It is like a symmetry, isn't it?
For a fixed position x, we would obtain a similar graph for y changes with time t, so the frequency we would hear will be $\frac{\Omega_1}{2\pi}$
Hmmm, brilliant, thanks

 

[haruspex]

In your notebook photo attached, replace x with t and K with $\Omega$. You see or hear sound of frequency $\Omega_1$ with amplitude or loudness changing with frequency $\Omega_2$.

To clarify, $\Omega_1$ would be the angular frequency, not the "heard frequency".

 

[rude man]

For (c), start with letting x=0 to simplify the math.

You will not hear a sound at frequency $\Omega_1$. You will hear 2 and only 2 frequencies, namely $\omega_1$ and $\omega_2.$

The "beat" at $\omega_1 - \omega_2$ is the difference in the magnitudes of the signals at frequencies $\omega_1$ and $\omega_2$ over time. It is not a separate frequency signal.

Say the frequencies are 1000 Hz and 1001 Hz. You don't hear a sound at (1001 + 1000)/2 = 1000.5 Hz. And you certainly don't hear a sound at (1001 - 1000)/2 = 0.5 Hz; your ear is limited to about 20 Hz at the low end.

If you hooked up a spectrum analyzer to a microphone you would see 2 and only 2 signals, one at $\omega_1$ and one at $\omega_2$, bobbing up and down at their respective frequencies.

 

[Bruce_Pipi121]

For (c), start with letting x=0 to simplify the math.

You will not hear a sound at frequency $\Omega_1$. You will hear 2 and only 2 frequencies, namely $\omega_1$ and $\omega_2.$

The "beat" at $\omega_1 - \omega_2$ is the difference in the magnitudes of the signals at frequencies $\omega_1$ and $\omega_2$ over time. It is not a separate frequency signal.

Say the frequencies are 1000 Hz and 1001 Hz. You don't hear a sound at (1001 + 1000)/2 = 1000.5 Hz. And you certainly don't hear a sound at (1001 - 1000)/2 = 0.5 Hz; your ear is limited to about 20 Hz at the low end.

If you hooked up a spectrum analyzer to a microphone you would see 2 and only 2 signals, one at $\omega_1$ and one at $\omega_2$, bobbing up and down at their respective frequencies.

This is an interesting idea. So do you mean that the sound we hear will automatically be "decomposed" into two sound waves with distinct frequencies? I think you are probably right, it is like our brain is embedded with a spectrum analyzer, isn't it?

 

[haruspex]

Say the frequencies are 1000 Hz and 1001 Hz. You don't hear a sound at (1001 + 1000)/2 = 1000.5 Hz. And you certainly don't hear a sound at (1001 - 1000)/2 = 0.5 Hz;

Depends what you mean by hearing. As a subjective process, you will hear a pitch of 1000.5Hz, with an amplitude that varies at 1Hz (not 0.5).

 

[rude man]

Right. What you hear is what is being sent, and that is two signals at slightly different frequencies. The "beat" is the difference in frequencies which alternately reinforce and cancel each other at the beat frequency rate.

You can't talk of one resultant frequency since you can't write $cos(\omega_1 t) + cos(\omega_2 t)$ as $Acos(w_3 t + \phi)$ (to be as general as possible).

Another view: you're probably familiar with Fourier series. Taking my 1000Hz and 1001Hz example again, you would get zero coefficients at the first harmonic (1Hz) and all the way to 999th harmonic. The 1000th and 1001st harmonics would give you the coefficients of your original expression, then all subsequent harmonics would also be zero.

Careful not to confuse this linear process with a non-linear one: if you multiplied $cos(\omega_1t)$ with $cos(\omega_2t)$ you would generate new signals at frequencies $| \omega_1 - \omega_2|$ and $(\omega_1 + \omega_2)$. Now you really have generated new 1Hz and 2001Hz signals. (This is called "mixing" (and in particular "suppressed-carrier amplitude modulation". If you do the trig you can see all that. for yourself.

If you have further interest I can give you a link to one of my blogs discussing this often-encountered confusion between "beats" and "mixing". Even my Resnick & Halliday fell for it!

 

[rude man]

Depends what you mean by hearing. As a subjective process, you will hear a pitch of 1000.5Hz, with an amplitude that varies at 1Hz (not 0.5).

Sorry, not so. See my post 8.

 

[haruspex]

You can't talk of one resultant frequency since you can't write $cos(\omega_1 t) + cos(\omega_2 t)$ as $Acos(w_3 t + \phi)$ (to be as general as possible).

But you can write it as $A(t)\cos(\omega_3 t + \phi)$, which we experience as a single frequency of varying amplitude. At least, I do.

 

[rude man]

But you can write it as $A(t)\cos(\omega_3 t + \phi)$, which we experience as a single frequency of varying amplitude. At least, I do.

The "coefficient" A(t) is not a constant so that eliminates all possibility of expressing the heard sound train as that of a single frequency.

It's true that the heard frequency is a mix of 1000 and 1001 Hz; by a "mix" I mean a jumble, not the "mixing" associated with a non-linear process. I don't want to get into a discussion as to what constitutes a "subjective process". I am saying the sound is not that of a single frequency 1000.5 Hz which is the all too commonly held belief. 1000.5Hz just never appears.

 

[haruspex]

The "coefficient" A(t) is not a constant so that eliminates all possibility of expressing the heard sound train as that of a single frequency.

AS I wrote, I am using "hearing" to refer to perception. Varying the amplitude at 1Hz or slower makes no difference to the perceived pitch.

 

[jbriggs444]

The model that I assume for the inner ear is that the cilia in the cochlea are damped oscillators that are stimulated by sound. Because they are damped oscillators, they are stimulated by frequencies not only at their resonant frequency but by frequencies near their resonant frequency.

So you have this array of oscillators whose resonant frequencies are, all over the place near 1000 Hz and you expose them to this mix of pure 1000 and 1001 Hz tones. They will all be stimulated to varying degrees with the peak stimulation centered at 1000.5 Hz. For very selective oscillators, one might note a bi-modal distribution. I do not believe that human ears perceive that audible effect in this case.

 

[haruspex]

The model that I assume for the inner ear is that the cilia in the cochlea are damped oscillators that are stimulated by sound. Because they are damped oscillators, they are stimulated by frequencies not only at their resonant frequency but by frequencies near their resonant frequency.

So you have this array of oscillators whose resonant frequencies are, all over the place near 1000 Hz and you expose them to this mix of pure 1000 and 1001 Hz tones. They will all be stimulated to varying degrees with the peak stimulation centered at 1000.5 Hz. For very selective oscillators, one might note a bi-modal distribution. I do not believe that human ears perceive that audible effect in this case.

Interesting model.
I plugged in the general solution for a forced damped oscillator and got that if a resonator resonate at $\omega$ and two incoming frequencies are $\omega_1 <\omega_2$ then peak power occurs when $\omega$ satisfies $\frac{\epsilon_1}{\epsilon_2}=(\frac{m^2\epsilon_1^2+b^2}{m^2\epsilon_2^2+b^2})^2$, where $\epsilon_1=(\frac{\omega}{\omega_1})^2-1$ and $\epsilon_2=1-(\frac{\omega}{\omega_2})^2$.
If the frequencies are all close together then we can drop the $\epsilon^2$ terms:
$\omega^2=\frac{2\omega_1^2\omega_2^2}{\omega_1^2+\omega_2^2}$