Question

The Shinkansen, the Japanese "bullet" train, runs at high speed from Tokyo to Nagoya. Riding on the

Answer 1

The speed of the Shinkansen is about 70 m/s ≈ 250 km/h

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Further explanation

Let's recall the Doppler Effect formula as follows:

$\large {\boxed {f' = \frac{v + v_o}{v - v_s} f}}$

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!

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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,

$f'_1 = \frac{v + v_o}{v - v_s} \times f_s$

$f'_1 = \frac{v + v_o}{v - 0} \times f_s$

$f = \frac{v + v_o}{v} \times f_s$ → Equation 1

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As you recede from the crossing,

$f'_2 = \frac{v - v_o}{v + v_s} \times f_s$

$f'_2 = \frac{v - v_o}{v + 0} \times f_s$

$\frac{2}{3}f = \frac{v - v_o}{v} \times f_s$ → Equation 2

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Next, we will solve the two equations above by dividing them

( Equation 1 ÷ Equation 2 ):

$f : \frac{2}{3}f = (\frac{v + v_o}{v} \times f_s) : (\frac{v - v_o}{v} \times f_s)$

$1 : \frac{2}{3} = (v + v_o) : (v - v_o)$

$3 : 2 = (v + v_o) : (v - v_o)$

$2(v + v_o) = 3(v - v_o)$

$2v + 2v_o = 3v - 3v_o$

$3v_o + 2v_o = 3v - 2v$

$5v_o = v$

$v_o = \frac{1}{5}v$

$v_o = \frac{1}{5}(343)$

$v_o \approx 70 \texttt{ m/s}$

$v_o \approx 250 \texttt{ km/h}$

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Learn more

• Doppler Effect : brainly.com/question/3841958
• Example of Doppler Effect : brainly.com/question/810552

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Answer details

Grade: College

Subject: Physics

Chapter: Sound Waves

Answer 2

Answer: 68.6 m/s

Explanation:

We have the following data:

$u$ is the speed of the train

$v=343 m/s$ is the speed of sound in air

$f$ is the frequency of the source of sound (the stationary signal)

$f'$ is the required frequency

Now, this can be expressed mathematically as follows:

When the train approaches the source of sound the frequency is $f$:

$f=(1+\frac{u}{v}) f'$ (1)

When the train recedes the source of sound the frequency is $\frac{2}{3}f$:

$\frac{2}{3}f=(1-\frac{u}{v}) f'$ (2)

Let's divide (1) by (2) to simplify and the find $u$:

$\frac{3}{2}=\frac{1+\frac{u}{v}}{1-\frac{u}{v}}$ (3)

Isolating $u$:

$u=\frac{v}{5}$ (4)

$u=\frac{343 m/s}{5}$ (5)

Finally:

$u=68.6 m/s$ This is the speed of the train