Two loudspeaker `M` and `N` are located `20m` apart and emit sound at frequencies `118Hz` and `121Hz`, respectively. A car is intially at a point `P`, `1800m` away from the midpoint `Q` of the line `MN` and moves towards `Q` constantly at `60 km//hr` along the perpendicular bisector of `MN`. It 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 `330ms^(-1)`. Which of the following statement (s) is (are) true regarding the sound heard by the person?

Options

A. The plot below represents schematically the variation of beat frequency with time
B. `v_(P)+v_(R)=2v_(Q)`
C. The plot below represents schematically the variation of beat frequency with time
D. The rate of change in beat frequency is maximum when the car passes through Q

Answer

A, B, D

Explanation

Sped of car, `V=60 km//h`
`(500)/(3) m//s`
At a point S,between P & Q
`v_m=v_m((c+v costheta)/(c)),v_n=v_n((v+v costheta)/(c))`
`implies trianglev=(v_n-v_n)(1-(v costheta)/(c))`
similarly between Q & R
`implies trianglev=(v_n-v_n)(1-(v costheta)/(c))`
`(d (trianglev))/dt=pm(v_n-v_n)v/csintheta(d theta)/dt`
`theta approx 0^(@)` at P & R as they are large distance apert
`implies` slope of graph is zero
at Q,`theta=90^(@)`
`implies sin theta` is max
also valued of `(d theta)/(dt)` is max
as `(d theta)/(dt)=v/x` where V is its velocity and r is length of kine joining P & S
and r is minimum at Q
`implies` slope is maximum at Q
`implies` (A) & (B) are correct
At p,`triangle v=(v_n-v_n)(1+(v)/(c)) ( theta=0^(@)`
At R, `triangle v=(v_n-v_n)(1-(v)/(c)) ( theta=0^(@)`
At Q, `triangle v=(v_n-v_n) (theta=90^(@))`
`implies trianglev_p+trianglev_R=2(v_n-v_n)=2trianglev_Q` Hence (B) is correct.