A police van,moving at 22 , m/s,chases a moto | vedclass

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(B) The apparent frequency $f'$ of the horn of the police van as heard by the motorcyclist is given by the Doppler effect formula:$f' = \left( \frac{v

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(B) The apparent frequency $f'$ of the horn of the police van as heard by the motorcyclist is given by the Doppler effect formula:$f' = \left( \frac{v - u_m}{v - v_s} ight) f_s$Given $v = 330 \, m/s$,$v_s = 22 \, m/s$,and $f_s = 176 \, Hz$:$f' = \left( \frac{330 - u_m}{330 - 22} ight) 	imes 176 = \left( \frac{330 - u_m}{308} ight) 	imes 176 \quad \dots (i)$The apparent frequency $f''$ of the stationary siren as heard by the motorcyclist is:$f'' = \left( \frac{v + u_m}{v} ight) f_{sireen}$Given $f_{sireen} = 165 \, Hz$:$f'' = \left( \frac{330 + u_m}{330} ight) 	imes 165 \quad \dots (ii)$Since the motorcyclist does not observe any beats,the frequencies must be equal $(f' = f'')$:$\left( \frac{330 - u_m}{308} ight) 	imes 176 = \left( \frac{330 + u_m}{330} ight) 	imes 165$$\frac{330 - u_m}{308} 	imes 176 = \frac{330 + u_m}{2}$$(330 - u_m) 	imes 0.5714 	imes 176 = (330 + u_m) 	imes 0.5$Simplifying the ratio: $\frac{176}{308} = \frac{4}{7}$ and $\frac{165}{330} = \frac{1}{2}$.$\frac{4(330 - u_m)}{7} = \frac{330 + u_m}{2}$$8(330 - u_m) = 7(330 + u_m)$$2640 - 8u_m = 2310 + 7u_m$$15u_m = 330$$u_m = 22 \, m/s$.

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