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ratelena [41]
3 years ago
10

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

Physics
2 answers:
Goryan [66]3 years ago
7 0

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

\texttt{ }

<h3>Further explanation</h3>

Let's recall the Doppler Effect formula as follows:

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

<em>f' = observed frequency</em>

<em>f = actual frequency</em>

<em>v = speed of sound waves</em>

<em>v_o = velocity of the observer</em>

<em>v_s = velocity of the source</em>

<em>Let's tackle the problem!</em>

\texttt{ }

<u>Given:</u>

initial observed frequency = f'_1 = f

final observed frequency = f'_2 = ²/₃ f

speed of sound in air = v = 343 m/s <em>(assumption)</em>

velocity of the source = v_s = 0 m/s

<u>Asked:</u>

velocity of the observer = v_o = ?

<u>Solution:</u>

<em>We will use the formula of </em><em>Doppler Effect.</em>

<em>As you approach the crossing,</em>

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

\texttt{ }

<em>As you recede from the crossing,</em>

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

\texttt{ }

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}

\texttt{ }

<h3>Learn more</h3>
  • Doppler Effect : brainly.com/question/3841958
  • Example of Doppler Effect : brainly.com/question/810552

\texttt{ }

<h3>Answer details</h3>

Grade: College

Subject: Physics

Chapter: Sound Waves

jeyben [28]3 years ago
6 0

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

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