Two loudspeakers M and N are located 20m apart and emit sound at frequencies 118Hz and 121Hz, respectively. A car is initially at point P, 1800m away from the point Q of the line MN and moves towards Q constantly at 60km/h along the perpendicular bisector of MN. If crosses Q and eventually reaches a point R, 1800m away from Q. Let V(t) represent the beat frequency measured by a person sitting in the car at time t. Let $V_{P}, V_{Q} and V_{R}$ be the beat frequencies measured at locations P, Q and R respectively. The speed of sound in air is 330m/s. Which of the following statement(s) is (are) true regarding the sound heard by the person?

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The plot below represents schematically the vibration of beat frequency with time

The rate of change in beat frequency is maximum when the car parallel through Q.

$V_{P} + V_{R} = 2 V_{Q}$

The plot below represents schematically the variation Of beat frequency with time.

answer is A, B, C.

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Detailed Solution

$V_{p} = 121 - 118 [\frac{V + V \cos θ}{V}] = 3 V_{R} = (121 - 118) [\frac{V - V_{c} \cos θ}{V}] v P + v R = 2 V Q$

A is correct

$V = 3 | \frac{V + V_{e} \cos α α}{V}] \frac{d V}{d α} = - 3 [\frac{V_{c} \sin α}{V}] | \frac{d V}{d α} |$ is

When $\sin α = 1$ $α = 90^{0}$

(b) is correct

$\frac{d V}{d t} = \frac{3 V_{c}}{V} (- \sin α) \frac{d α}{d t}$ $\tan α = \frac{10}{x}$

$\sec^{2} α \frac{d α}{d t} = \frac{- 10}{x^{2}} \frac{d x}{d t} \frac{d α}{d t} = \frac{- 10 V}{x^{2} \sec^{2} α} \frac{d V}{d t} = \frac{- 3 V_{c}}{V} \sin x \times [\frac{- 10 V}{x^{2} \sec^{2} α}] = \frac{30 V_{c} \sin α}{x^{2} \sec^{2} α} A t α = 90^{0} \frac{d V}{d t} = \max$