Answer: 68.6 m/s
Explanation:
We have the following data:
[tex]u[/tex] is the speed of the train
[tex]v=343 m/s[/tex] is the speed of sound in air
[tex]f[/tex] is the frequency of the source of sound (the stationary signal)
[tex]f'[/tex] is the required frequency
Now, this can be expressed mathematically as follows:
When the train approaches the source of sound the frequency is [tex]f[/tex]:
[tex]f=(1+\frac{u}{v}) f'[/tex] (1)
When the train recedes the source of sound the frequency is [tex]\frac{2}{3}f[/tex]:
[tex]\frac{2}{3}f=(1-\frac{u}{v}) f'[/tex] (2)
Let's divide (1) by (2) to simplify and the find [tex]u[/tex]:
[tex]\frac{3}{2}=\frac{1+\frac{u}{v}}{1-\frac{u}{v}}[/tex] (3)
Isolating [tex]u[/tex]:
[tex]u=\frac{v}{5}[/tex] (4)
[tex]u=\frac{343 m/s}{5}[/tex] (5)
Finally:
[tex]u=68.6 m/s[/tex] This is the speed of the train
The speed of the Shinkansen is about 70 m/s ≈ 250 km/h
[tex]\texttt{ }[/tex]
Let's recall the Doppler Effect formula as follows:
[tex]\large {\boxed {f' = \frac{v + v_o}{v - v_s} f}}[/tex]
f' = observed frequency
f = actual frequency
v = speed of sound waves
v_o = velocity of the observer
v_s = velocity of the source
Let's tackle the problem!
[tex]\texttt{ }[/tex]
Given:
initial observed frequency = f'_1 = f
final observed frequency = f'_2 = ²/₃ f
speed of sound in air = v = 343 m/s (assumption)
velocity of the source = v_s = 0 m/s
Asked:
velocity of the observer = v_o = ?
Solution:
We will use the formula of Doppler Effect.
As you approach the crossing,
[tex]f'_1 = \frac{v + v_o}{v - v_s} \times f_s[/tex]
[tex]f'_1 = \frac{v + v_o}{v - 0} \times f_s[/tex]
[tex]f = \frac{v + v_o}{v} \times f_s[/tex] → Equation 1
[tex]\texttt{ }[/tex]
As you recede from the crossing,
[tex]f'_2 = \frac{v - v_o}{v + v_s} \times f_s[/tex]
[tex]f'_2 = \frac{v - v_o}{v + 0} \times f_s[/tex]
[tex]\frac{2}{3}f = \frac{v - v_o}{v} \times f_s[/tex] → Equation 2
[tex]\texttt{ }[/tex]
Next, we will solve the two equations above by dividing them
( Equation 1 ÷ Equation 2 ):
[tex]f : \frac{2}{3}f = (\frac{v + v_o}{v} \times f_s) : (\frac{v - v_o}{v} \times f_s)[/tex]
[tex]1 : \frac{2}{3} = (v + v_o) : (v - v_o)[/tex]
[tex]3 : 2 = (v + v_o) : (v - v_o)[/tex]
[tex]2(v + v_o) = 3(v - v_o)[/tex]
[tex]2v + 2v_o = 3v - 3v_o[/tex]
[tex]3v_o + 2v_o = 3v - 2v[/tex]
[tex]5v_o = v[/tex]
[tex]v_o = \frac{1}{5}v[/tex]
[tex]v_o = \frac{1}{5}(343)[/tex]
[tex]v_o \approx 70 \texttt{ m/s}[/tex]
[tex]v_o \approx 250 \texttt{ km/h}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: College
Subject: Physics
Chapter: Sound Waves
