Question

A source of sound and detector are arranged as shown in Fig. The detector is moving along a circle with constant angular speed `omega`. It start from the shown location in anticlockwise direction at `t=0`
(Take velocity of sound in air as v)
Q. Find the time at which the detector will hear the maximum frequency for the first time.
A. `(pi)/((3omega))`
B. `(5pi)/((3omega))`
C. `(4pi)/((3omega))`
D. `(pi)/(omega)`

Answer 1

Correct Answer - B

The instant at which the detector hears maximum and minimum frequencies are shown in Fig. A corresponds to minimum frequency while B corresponds to maximum frequency
`costheta=(R )/(2R)=(1)/(2)`
`impliestheta=(pi)/(3)`
Let the detector reaches A in time `t_0` (measured from the starting instant). then
`theta=omegat_0`
`implies(pi)/(3)=omegat_0`
`impliest_0=(pi)/(3omega)`
let the detector reaches B in time `t_(01)` (measured from A). Then,
`2pi-2theta=omegaxxt_(01)`
`impliest_(01)=(4pi)/(3omega)`
The time taken after which the detector hears the maximum frequency for the first time is
`t_0+t_(01)=(5pi)/(3omega)`
The interval between maximum and minimum frequencies as received by the detector for the first time is
`t_(01)=(4pi)/(3omega)`