Physics, asked by sana999, 1 year ago

Explain Doppler effect.

Answers

Answered by Anonymous
3

When wave energy like sound or radio waves travels from two objects, the wavelength can seem to be changed if one or both of them are moving. This is called the Doppler effect.


The Doppler effect causes the received frequency of a source (how it is perceived when it gets to its destination) to differ from the sent frequency if there is motion that is increasing or decreasing the distance between the source and the receiver. This effect is readily observable as variation in the pitch of sound between a moving source and a stationary observer. Imagine the sound a race car makes as it rushes by, whining high pitched and then suddenly lower. Vrrrm-VROOM. The high pitched whine is caused by the sound waves being compacted as the car approaches you, the lower pitched VROOM comes after it passes you and is speeding away. The waves are spread out.



Answered by fanbruhh
2
\huge{hey }

\huge{ \mathfrak{here \: is \: answer}}

DOPPLER'S EFFECT
Non - Relativistic Treatment
As most readers must have read in physics according to. This phenomenon the pitch sound observed by an observer is different in two cases, firstly when both the source and observer are stationary and secondly when there is relative motion between them. When source and observer are approaching each other, the apparent frequency is increased and when receding, decreased. This phenomenon is called doppler's effect which occurs with all kinds of wave motion, although in case of mechanical waves involving a material medium, its nature is some what different from the case of electromagnetic waves such as light where no medium is involved . In the case of light waves, if spectrum is observed in a spectrometer, a shift in the position of spectral line from the original position is observed with the motion of source.

we first consider the case when the observer is at rest and the source is moving towards it velocity u. let the frequency of the source be f so that the time period
T=1/f
and the wave speed be c in a time time interval T during which one cycle of wave is emitted the wave progresses a at distance cT but in the same period the source wave in the same direction distance u T hence wavelength which is distance between two successive maximum in the wave is cT-uT instead of cT.
Hence the corresponding frequency denoted by f1' its given by

f1'= c/ T(c-u)= cf/c-u

= f/ 1-(c/u)........ (1)

which represents and Apparent increase in the frequency if the source is moving away from the observer we put - u in place of u in equation( 1st) and get the corresponding f1'

in the second case When the source is stationary but the observer is moving with the velocity v from the source the wave is speed relative to the observer is not c but c-v and hence apparent frequency in the case is.

f \tiny{2} = \frac{c - v}{ \lambda} 
= c-v/c/f
= c-v/c*f

1 - \frac{v}{c} f \: .......(2)

now both observer and source are moving

\bf{\lambda \tiny{1} = \frac{c}{f \tiny{1}} }<br />= \: \frac{c - u}{ \lambda}

\huge \boxed { \boxed{ \boxed{hope \: it \: helps}}}

\huge{ \mathfrak{thanks}}
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