Question Number 35879 by NECx last updated on 25/May/18
$${Find}\:{the}\:{velocity}\:{of}\:{the}\:{source}\: \\ $$$${along}\:{the}\:{line}\:{joining}\:{the}\:{source} \\ $$$${to}\:{a}\:{stationary}\:{observer}\:{if},{as}\:{a} \\ $$$${result}\:{of}\:{the}\:{motion},{the}\:{frequency} \\ $$$${of}\:{the}\:{note}\:{heard}\:{is} \\ $$$$\left.{a}\right){increased}\:{in}\:{the}\:{ratio}\:\mathrm{18}:\mathrm{15} \\ $$$$\left.{b}\right){decreased}\:{in}\:{the}\:{ratio}\:\mathrm{15}:\mathrm{18} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/May/18
$$\left.{a}\right)\frac{{f}'}{{f}}=\frac{{v}}{{v}−{v}_{{s}} }=\frac{\mathrm{18}}{\mathrm{15}} \\ $$$$\mathrm{18}{v}−\mathrm{18}{v}_{{s}} =\mathrm{15}{v} \\ $$$$\mathrm{3}{v}=\mathrm{18}{v}_{{s}\:\:\:} {v}_{{s}} =\frac{{v}}{\mathrm{6}}\:\:\:{v}_{{s}} =\frac{\mathrm{330}}{\mathrm{6}}=\mathrm{55}{meter}/{second} \\ $$$$\left.{b}\right)\frac{{f}'}{{f}}=\frac{{v}}{{v}+{v}_{{s}} }=\frac{\mathrm{15}}{\mathrm{18}} \\ $$$$\mathrm{15}{v}+\mathrm{15}{v}_{{s}} =\mathrm{18}{v} \\ $$$${v}_{{s}} =\frac{\mathrm{3}{v}}{\mathrm{15}}=\frac{{v}}{\mathrm{5}} \\ $$$${v}_{{s}} =\frac{\mathrm{330}}{\mathrm{5}}=\mathrm{66}{m}/{sec} \\ $$