Stepping to the beat

Because of Doppler Effect, there is a difference, which is $2$ Hz, between the frequencies you hear from the two trumpet players.

We will use the following equations:

1) For sounds whose source is approaching the observer, $f_{observed} = [\frac{v_{sound}}{v_{sound}-v_{source}}] \times f_{source}$

, and

2) For sounds whose source is moving away from the observer, $f_{observed} = [\frac{v_{sound}}{v_{sound}+v_{source}}] \times f_{source}$

, where $f_{observed}$ is the frequency of the sound heard, $f_{source}$ is the frequency of the sound at the source, $v_{sound}$ is the speed of sound, which is $340 m/s$ .

Now, we set up the equation: $[(\frac{340}{340-v})440] - [(\frac{340}{340+v})440] = 2$

,which can be simplified into:

$v^2+149600v-115600=0$ .

The positive solution of the quadratic above is $v=0.773m/s$ , which is the desired answer.

let the velocity of players be v then f1 - f2 =2=440 {(340/340-v)-(340/340+v)} Applying binomial approx., v=.773m/s

first trumpet player is moving toward us , so the equation can be stated as following

frequency heard 1 = [340/(340-v) ]*440hz

second trumpet player is moving away from us , so the equation can be stated as following

frequency heard 2 = [340/(340+v) ]*440hz

since we hear a beat with frequency of 2 hz, therefore

frequency heard 1 - frequency heard 2 = 2 Hz

[ 340/(340-v) - [340/(340+v) ] *440 = 2

v^2 + 149600v -115600 = 0

Roots :

V1 = -1.496007727e+5

V2 = 0.7727232814

since we assume that both trumpet players are moving to the right, therefore, we take the second root ( + )

0.772

as our answer

-JA-

The perceived frequency than the player that comes close with velocity v is: f=(340/(340-v))440.

The perceived frequency than the player that gets away with velocity v (we can absume the same of the first player) is: f'=(340/(340+v))440.

So: f - f' = (340/(340-v))440 - (340/(340+v))440 = 2 Hz

So: v^2-340*440v-340^2=0

and the little root v = 0.772 m/s is the right root.

Let the speed of trumpet players be v, then applying the resultant frequency equation as it is clear that one trumpet player is marching away from the observer (me) and other trumpet player is marching towards me gives, 440/(1 + (340/v)) - 440/(1-(340/v)) = 2 which on solving further gives v=0.772 m/s.We have to neglect the negative root of the quadratic equation as speed can never be negative.

assume the speed of the trumpet players to be $v$ .

apparent frequency heard due to the trumpet player moving away is

$\frac{440(340)}{340+v}$ .

apparent frequency heard due to the trumpet player moving towards me is

$\frac{440(340)}{340-v}$ .

we just need to solve

$\frac{440(340)}{340-v} - \frac{440(340)}{340+v} = 2$

the answer comes out to be $0.77272$ in MKS units.

use Doppler effect

$f_{b} = f_{o_{1}} - f_{o_{2}} = 2$

$(v / (v - v_{s})) f_{s} - ((v / (v + v_{s})) f_{s} = 2$

which $v =$ 340 m/s; $f_{s} = 440 Hz$

then we'll have :

$v_{s}^{2} + 340 . 440 v_{s} - 340^{2} = 0$

finally we'll have :

$v_{s} = 0.7727$ m/s

Well, the beat frequency is expressed as $f = f_{1} - f_{2}$ being $f_{1}$ and $f_{2}$ the frequencies that are arriving to you. Suppose $f_{1}$ is the frequency of the sound that you receive from the person who is comming closer to you. Then $f_{1} = f_{0} \frac{v}{v - V}$ According to Doppler Effect, where $f_{0}$ is the frequency that the marching person hears from its own instrument, $v$ is the speed of sound and $V$ is the velocity of the marching person. Then $f_{2}$ must be the frequency you are receiving from the person who is going further from you. So $f_{2} = f_{0} \frac{v}{v + V}$ Then $f$ may be expressed as $f = f_{0} v \frac{2V}{v^{2} - V^{2}}$ The solution for $V$ is easily given by Bhaskara: $V = \frac{- f_{0} v + v \sqrt{f_{0}^{2} + f^{2}}}{f} \approx 0.773 m/s$

Let, v = velocity of sound (340 m/s)

u = velocity of the source (trumpet players)

f = 440 Hz

f' - f'' =2

f'= vf/(v - u),

f'' = vf/(v + u),

vf/(v - u) - vf/(v + u) = 2

u^2 + 149,600 u – 115,600 = 0

u = 0.772723281 m/s

Doppler effect