The drawing shows three situations-A, B, and C-in which an observer and a source

Question

The drawing shows three situations-A, B, and C-in which an observer and a source of electromagnetic waves are moving along the same line. In each case the source emits a wave that has a frequency of 4.57 x 1014 Hz. The arrows in each situation denote velocity vectors of the observer and source relative to the ground and have the magnitudes indicated (v or 2v), where the speed v is 1.65 x 106 m/s. Calculate the observed frequency in each of the three cases. Source 2u -Select- -Select-?? -Select-? C: fo=

Explanation / Answer

This is the problem of Doppler effect.
Here electromagnetic wave causes
the effect in place of sound and
the velocity of wave is c=3*108 m/s.
So by applying the Doppler’s formula

(a) f=(c+v)f0/(c+2v)=(1+v/c)f0/(1+2v/c)
=(1+1.65/300)*4.57*1014/(1+3.3/300)
=4.55*1014 Hz

(b)f=(c?v)f0/(c+2v)?4.495*1014 Hz

(c)f=(c+v)f0/(c?2v)?4.65*1014 Hz

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