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Ganezh [65]
2 years ago
5

A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red

light of wavelength 650nm appear green to him, with a wavelength of 520nm . The police officer writes out a traffic citation for speeding. How fast was the physicist traveling, according to his own testimony?
Physics
1 answer:
Orlov [11]2 years ago
5 0

The physicist traveling, according to his own testimony at -6.6 × 10⁷ m/s.

<h3>How fast was the physicist traveling, according to his own testimony?</h3>

Using the formula for doppler shift for light,

λ' = λ√[(1 + v/c)/(1 - v/c)] where

  • λ = wavelength of source,
  • λ' = wavelength of observer,
  • v = speed of source and
  • c = speed of light

Given that the driver is moving away from the stop light, we take the driver as the source. Since, the Doppler shift made the red light of wavelength 650nm appear green to him, with a wavelength of 520nm. we have

  • λ' = wavelength of source = 650 nm,
  • λ' = wavelength of observer = 520 nm

So, substituting the values of the variables into the equation, we have

λ' = λ√[(1 + v/c)/(1 - v/c)]

520 nm = 650 nm√[(1 + v/c)/(1 - v/c)]

520/650 = √[(1 + v/c)/(1 - v/c)]

0.8 = √[(1 + v/c)/(1 - v/c)]

Squaring both sides, we have

0.8² = (1 + v/c)/(1 - v/c)

0.64 = (1 + v/c)/(1 - v/c)

0.64(1 - v/c) = (1 + v/c)

0.64 - 0.64v/c = 1 + v/c

0.64 - 1 = v/c + 0.64v/c

-0.36 = 1.64v/c

-0.2195 = v/c

v = -0.22c

v = -0.22 × 3 × 10⁸ m/s

v = -0.66 × 10⁸ m/s

v = -6.6 × 10⁷ m/s

So, the physicist traveling, according to his own testimony at -6.6 × 10⁷ m/s.

Learn more about doppler shift for light here:

brainly.com/question/28499579

#SPJ1

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A)i) 1. constant,  2. constant,  3. constant,  4. decrease

   ii)  frecuency increase

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B)   L_b = 0.534 m

Explanation:

We can approximate the violin string as a system of a fixed string at its two ends, therefore we have a node at each end and a maximum in the central part for a fundamental vibration,

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where n is an integer

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and the speed of the wave is given by

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A)

i) In this case the woman decreases the length of the rope L = L₂

      therefore the wavelength changes

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as L₂ <L₀ the wavelength is

         λ₂ < λ₀

The tension of the string is given by the force of the plug as it has not moved, the tension must not change and the density of the string is a constant that does not depend on the length of the string, therefore the speed of the string wave in the string should not change.

ii) how we analyze if the speed of the wave does not change

         v = λ  f

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     fy> fx

iii) It is asked to find the length of the chord

let's use the initial equations

            λ  = 2L / n

            v = λ  f

            v = 2L / n f

            v = √ T /μ

we substitute

           2 L / n f = √ T /μ

           L = n /2f   √T/μ

this is the length the string should be for each resonance

b) in this part they ask to calculate the frequency

         f = n / 2L √ T /μ

the linear density is

         μ = m / L

         μ = 2.00 10⁻³ / 60.0 10⁻²

         μ = 3.33 10⁻³ kg / m

we assume that the length is adequate to produce a fundamental frequency in each case

f_{a} = 440Hz

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        λ = 2 0.60 / 1

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        v = λ f

        v = 1.20 440

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        v² = T /μ

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       T = 528² 3.33 10⁻³

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Let's find the length of the chord for fb

f_{b} = 494 hz

        L_b = 1 /(2 494)  √(9.28 10² / 3.33 10⁻³)

        L_b = 0.534 m

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