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Obtain the equation of frequency observed by moving observer and stationary source. |
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Answer» Solution :When the observe is MOVING with velocity `v _(0)` towards the source and the source is at rest, we have to proceed in a different manner. We work in the reference frame of the moving observer. In this reference frame the source and medium are approaching at speed `v _(0)` and the speed withwhich the wave approaches is `v _(0)+v .` Following a similar PROCEDURE as in the previous case, we find that the time interval between the arrival of the FIRST and the `(N+1) ^(th)` crests is, `t _(n+1) -t_(1) = n T_(0) -(n v _(0) T_(0))/(v _(0) +v)` To observer thus, measures the period of hte wave to be, ` T = (t _(n +1) -t _(1))/(n)` `T_(0) (1- (v _(0))/(v _(0) + v)) = T _(0) ((v)/(v _(0) + v ))` `= T _(0) ((1)/(v _(0) //v + 1))` `= T _(0) (1 + (v _(0))/( v))` giving, `v = v _(0) [ 1 + (v _(0))/(v)]` When observer moves toward stationary source then observed FREQUENCY is less than original frequency. When observe moves away from stationary source then `v _(0)` should be used in place of `-v _(0)` in equation `therefore` Frequency observed by observer, `v =v _(0) [1- (v _(0))/(v)]` Thus, frequency observed by observer will be greater than original frequency. |
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