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(B) First,convert the velocities from $km/h$ to $m/s$: $v_{S1} = 108 	imes \frac{5}{18} = 30 \ m/s$ $v_O = 36 	imes \frac{5}{18} = 10 \ m/s$ For obs

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$8$

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(B) First,convert the velocities from $km/h$ to $m/s$: $v_{S1} = 108 	imes \frac{5}{18} = 30 \ m/s$ $v_O = 36 	imes \frac{5}{18} = 10 \ m/s$ For observer $O$ and source $S_2$: The source $S_2$ is stationary $(v_s = 0)$. The observer $O$ moves towards $S_2$ with a velocity component along the line joining them. Since $O$ moves directly towards $S_2$,the velocity of the observer is $v_o = 10 \ m/s$. Using the Doppler effect formula: $f_2 = f_0 \left( \frac{v + v_o}{v} ight) = 120 \left( \frac{330 + 10}{330} ight) = 120 \left( \frac{340}{330} ight) \approx 123.64 \ Hz$. For observer $O$ and source $S_1$: The distance between $S_1$ and $S_2$ is $800 \ m$,and $O$ is $600 \ m$ from $S_2$. The distance $OS_1 = \sqrt{800^2 + 600^2} = 1000 \ m$. The angle $	heta$ such that $\cos 	heta = \frac{800}{1000} = 0.8$ and $\sin 	heta = \frac{600}{1000} = 0.6$. The velocity component of $S_1$ towards $O$ is $v_{s1} = v_{S1} \cos 	heta = 30 	imes 0.8 = 24 \ m/s$. The velocity component of $O$ towards $S_1$ is $v_{o1} = v_O \sin 	heta = 10 	imes 0.6 = 6 \ m/s$. Using the Doppler effect formula: $f_1 = f_0 \left( \frac{v + v_{o1}}{v - v_{s1}} ight) = 120 \left( \frac{330 + 6}{330 - 24} ight) = 120 \left( \frac{336}{306} ight) \approx 131.76 \ Hz$. Beat frequency $= |f_1 - f_2| = 131.76 - 123.64 = 8.12 \ Hz$. The number of beats heard is approximately $8$.

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