Question

You are standing on a train station platform as a train goes by close to you. As the train approaches, you hear the whistle sound at a frequency of f1 = 98 Hz. As the train recedes, you hear the whistle sound at a frequency of f2 = 76 Hz. Take the speed of sound in air to be v = 340 m/s(a) Find an equation for the speed of the sound source v., in this case it is the speed of the train. Express your answer in terms of f2. and v.

Answer 1

Answer: v = (33320 - 340f2)/( f2 + 98)

Explanation: this question refers to Doppler effect, hence,

f1 = { V / (V - v)} x f2

f1 = 98Hz, f2 = 76 Hz, V = 340m/s, v = ?, fs = fsourcs

For f1,

98 = { 340 / (340 + v)} x fs ...(i)

for f2,

f2 = { 340 / (340 - v)} x fs ... (ii)

Divide (ii) by (i)

98/f2 = [{ 340 / (340 + v)} x fs]÷ [ 340 / (340 - v)} x fs]

98/f2 = {340 / (340 + v)} x fs x (340 - v)} / 340 x fs

98/f2 = (340 + v)/ (340 - v)

Cross multiplying

98(340 - v) = f2 (340 + v)

33320 - 98v = 340f2 + vf2

Collecting like terms

vf2 +98v = 33320 - 340f2

v(f2 +98) = 33320 - 340f2

v = (33320 - 340f2)/( f2 + 98)

Answer 2

The dopler effect allows finding the result for the speed of the train as a function of the frequency of when it moves away and the speed of sound is:

• The expression for the speed of the train is: $v_s = v \ \frac{98-f_2}{98 + f_2}$  

• The velocity for this case is: vs = 43 m / s

The Dopler effect is the change in the frequency of the sound due to the relative movement of the source and the observer.

In this case the observer is standing on the platform whereby the movement is from the sound source.

             $f' = f \ \frac{v}{v \mp v_s}$

Where f’ is the perceived frequency, f the emitted frequency, v the speed of sound and $v_s$  the speed of the source, the negative sign corresponds when they are approaching and the positive sign when they are moving away.

Indicate that when the train approaches the perceived frequency is f₁ = 98 Hz and when the train moves away the frequency is f₂ = 76 Hz, the speed of sound is v = 340 m / s.

            $f_1 = f \ \frac{v}{v-v_s} \\f_2 = f \ \frac{v}{v+v_s}$  

We have a system of two equation with two unknowns, we can solve it.

        $f_1 ( v-v_s) = vf \\f_2( v + v_s) = vf$  

let's resolve.

        $f_1 (v-v_s) = f_2 ( v+v_s )\\v ( f_1-f_2) = v_s ( f1+f2) \\\\v_s = v \ \frac{f_1 - f_2}{ f_1+f_2}$

They indicate that the result is given as a function of f₂ and the speed of sound, let's substitute.

        $v_s = v \ \frac{98 - f_2}{98+ f_2}$  

Let's calculate the speed of the train.

        $v_s = 340 \ \frac{98-76}{98+76}$  

        $v_s$ = 43 m / s

In conclusion using the dopler effect we can find the result for the speed of the train as a function of the frequency of when it moves away and the speed of sound is:

• The expression for the speed of the train is: $v_s = v \ \frac{98-f_2}{98+f_2}$  

• The velocity for this case is: $v_s$  = 43 m / s

Learn more about the Doppler effect here:  brainly.com/question/5871604