# Doppler Effect

Discussion in 'Physics' started by logearav, Sep 26, 2011.

1. ### logearav Thread Starter Member

Aug 19, 2011
248
0
Revered members,
I have attached the image of Doppler effect explanation
I have the following doubts
1)t1 = L/v
2) t2 = T0 +( L + vsT0)/v
But why T0 comes here, instead of ( L + vsT0)/v

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2. ### BillO Distinguished Member

Nov 24, 2008
985
136
Because the next wave peak does not get emitted until To (one period of the wave) has passed. Then to that we need to add the time of propagation, ( L + vsT0)/v, so that t2 (the next time you detect the peak) = T0 + ( L + vsT0)/v.

3. ### logearav Thread Starter Member

Aug 19, 2011
248
0
Sir, I can't understand.

4. ### BillO Distinguished Member

Nov 24, 2008
985
136
Read 15.8.1 very carefully. It is well explained there.

T0 is the period of time between peaks emitted by the moving source, so some multiple of T0 must be a part of the time the observer detects between each successive peak after t1, as well as the time of propagation from the point of emission to the observer.

So, we have the following sequence;

$
t_{1}\ =\ L/v
t_{2}\ =\ T_{0}\ +\ (L\ +\ v_{s}T_{0})/v
t_{3}\ =\ 2T_{0}\ +\ (L\ +\ 2v_{s}T_{0})/v
t_{4}\ =\ 3T_{0}\ +\ (L\ +\ 3v_{s}T_{0})/v
.
.
.
t_{n+1}\ =\ (n)T_{0}\ +\ (L\ +\ (n)v_{s}T_{0})/v
.
.
.
$

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