# two source of sound are producing wave of frequency V1 and V2 where (V1-V2) is small show mathematically that the beat frequency is (v1-V2)

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### 1 Answer

Suppose one has two audible oscillations having the same amplitude with angular frequencies `omega_1` and `omega_2` :

`y_1(t) = A*sin(omega_1*t)` and `y_2 =A*sin(omega_2*t)`

Now suppose one adds together the oscillations to obtain the resultant:

`y_("tot")(t) =y_1(t)+y_2(t)=A(sin(omega_1*t)+sin(omega_2*t))` (1)

Remember the sum and difference of two angles` ` `a` and `b` :

`sin(a+b) =sin(a)*cos(b) +sin(b)*cos(a)`

`sin(a-b) =sin(a)cos(b) -sin(b)*cos(a)`

If one take `a+b = omega_1*t` and `a-b =omega_2*t` one obtains from (1):

`y_("tot") =2A[sin((omega_1+omega_2)/2*t)*cos((omega_1-omega_2)/2*t)]` (2)

What one person hear is the intensity of the sound and the intensity is a power mediated over a surface

` ``I =P/S =E/(S*t)` .

But we know that the power (energy/time) of an oscillator is proportional to the square of the oscillator amplitude.

`I = C*y_("tot")^2` , where `C` is a constant.

Therefore from (2) a person will hear the following intensity of sound:

`I=C*cos^2((omega_1-omega_2)/2*t)*sin^2((omega_1+omega_2)/2*t)` (3)

Now, if the difference `omega_1-omega_2` is small, one can write the Taylor series expansion corresponding to the function `cos^2(x)` for `x ` small around `x =0` .

`cos^2(x) =1-2x+...`

(upper terms can be neglected because `x` is small)

Hence the ` `expression (3) will become

`I =C(1-(omega_1-omega_2)*t)*sin^2((omega_1+omega_2)/2*t)`

**Thus, around `t=0` the modulating amplitude of the sound heard will be `(omega_1-omega_2)*t` or in other words equal to the difference of the frequencies of original sounds.**