Modern Physics for Scientists and Engineers
Modern Physics for Scientists and Engineers
4th Edition
ISBN: 9781133103721
Author: Stephen T. Thornton, Andrew Rex
Publisher: Cengage Learning
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Chapter 2, Problem 39P
To determine

Frequency of the signal received in beginning (f) and end (f) of the trip and time period of Frank and Marry in receiving these frequencies.

Expert Solution & Answer
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Answer to Problem 39P

frequency of the signal received in beginning is f=f0(1β)(1+β) and end of the trip is f=f0(1+β)(1β) and time period of Frank in starting is t1=Lv+Lc and at end of the trip is t2=LvLc and for Marry in starting time period is t1=Lγv and at end of the trip is t2=Lγv, in receiving these frequencies.

Explanation of Solution

Initially the signal sent by Marry will be received by the Frank. The source emitting n waves in a time interval T, here Marry going away with velocity (v) from frank that is source is moving away from observer

Write the equation to get the distance travelled by a signal. Thus,

d=cT+vT                                                                                                                (I)

Write the equation to get the wavelength of waves emitting n waves in a time interval T and travelling a distance d in that time interval. Thus,

λ=cT+vTn                                                                                                              (II)

Write the equation to show the relation between frequency and wavelength. Thus,

f=cλ

Where, c is the speed of the light, f is the frequency of the waves, and λ is the wavelength of the waves.

Substitute the value of wavelength from equation (II) in above equation. Thus,

f=ccT+vTn=cncT+vT                                                                                                                  (III)

Write the equation to get the total number of the signals in time interval T0. Thus,

n=f0T0                                                                                                                     (IV)

Where, T0=Tγ, T is the time measured in the observer’s (Frank) clock and γ=11v2c2 is the relativistic factor.

Substituting the value of n from equation (IV) in equation (III). Thus

f=cf0T0 cT+vT                                                                                                       

Substitute the T0=Tγ and 1v2c2=1γ in above equation

f=cf0(Tγ) cT+vT=1v2c2(cf0 c+v)=1v2c2(f0 1+vc)=1β2(f0 1+β)                                                                                                       

Because β=vc, simplify the above equation, thus,

f=(1β)(1+β)(f0 (1+β)2)f=f0(1β)(1+β)                                                                                                         (V)

Initially Marry and Frank receives same frequency.

At the end of the trip Marry going away from Frank thus, β changes to β in the above expression. Thus,

f=f0(1(β))(1+(β))                                                                                          =f0(1+β)(1β)        (VI)

Thus, marry initially receives signals at a rate of f with t1=Lγv and at end of the trip signals will be received at a rate of  f for t2=Lγv.

And, Frank initially receives signals at a rate of f with t1=Lv+Lc and at end of the trip signals will be received at a rate of f for t2=LvLc.

Where, L is the distance of the Marry’s Journey from earth to star

Conclusion:

Therefore, frequency of the signal received in beginning is f=f0(1β)(1+β) and end of the trip is f=f0(1+β)(1β) and time period of Frank in starting is t1=Lv+Lc and at end of the trip is t2=LvLc and for Marry in starting time period is t1=Lγv and at end of the trip is t2=Lγv, in receiving these frequencies

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Chapter 2 Solutions

Modern Physics for Scientists and Engineers

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