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Obtain an expression for the apparent frequency of sound for a sourcemoving away and towards a stationary observer. |
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Answer» Solution :Let `S_(1)` be the position of the source at t= 0. Let L be the distance `b//w`the observed and the source at S, Let v be the velocity of sound in air. Time taken for sound ( say crest ) to REACH the observer `t_(1) =( L)/( v)`......(1) Let the source travel further from` S_(1) `,and OCCUPY the position `S_(2)` AWAY from a stationary observer. distance `S_(1)S_(2) = v_(s) T_(0)` For a crest at `S_(2)` to reach the observer, it takes the time `t_(2) = T_(0) + (L+v_(s) T_(0))/( v)`....(2) At time nT, the source emits its `(n+1)^(TH)` crest and this reaches the observer at time. `t_(n+1) = n T_(0) + ( L+ v_(s) ( nT_(0)))/(v_(0))` Time of observation of n crests by an observer will be nT. `nT = ( nT_(0) + ( L + nv_(s) T_(0))/(v))-((L)/( v))` Time for I crest = `( nT_(0) + ( Lnv_(s) T_(0))/( v) - ( L)/( v)) + n ` `T = ( nvT_(0)+L+nv_(s) T_(0)-L)/(nv)` i.e, `T= T_(0)+ (v_(s) T_(0))/( v)`  or `T = T_(0) ((v+ v_(s))/(v))` Apparent frequency of soundheard by a stationary observer for a source of sound moving away from him. `f =f_(0)((v)/(v+v_(s)))` and apparent wavelength `lambda =( v+ v_(s))/(f_(0))` , change ( increase ) in wavelength. `Deltalambda = ( v+ v_(s))/( f_(0))- ( v)/(f_(0)) = ((1)/( f_(0))) ( v+ v_(s)-v)= (v_(s)) /(f_(0))` Percentage increase in `lambda = ( Delta lambda)/( lambda ) xx 100 = ( v_(s))/( f_(0) lambda)xx100` For a source moving TOWARDS the observer , change `v_(s)` to `- v_(s)` Then `f= f_(0) ((v)/(v-v_(s)))` Apparentwavelength `=( v- v_(s))/( f_(0))`, here `lambda lt lambda_(0)` |
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