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The frequency changes by `10%` as the source approaches a stationary observer with constant speed `v_S`, What sould be the percentage change in ferquency as the source recedes from the observer with the same speed? Given that `v_Sltltv` (v speed of sound in air).A. `14.3%`B. `20%`C. `16.7%`D. `10%` |
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Answer» Correct Answer - D When the source approaches the observer `f_1=f((v)/(v-v_s))=f(1-(v_s)/(v))^(-)=f(1+(v_s)/(v))` or `((f_1-f)/(f))xx100=(v_s)/(v)xx100=10` In the second case, when the source recedes from the observer `f_2=f((v)/(v+v_s))=f(1+(v_s)/(v))^(-)=f(1-(v_s)/(v))` `((f_2-f)/(f))xx100=-(v_s)/(v)xx100=-10` In the first case, observed frequency increases by `10%` while in the second case, observed frequency decreases by `10%`. |
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