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प्रश्न
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
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उत्तर १
Ultrasonic beep frequency emitted by the bat, ν = 40 kHz
Velocity of the bat, vb = 0.03 v
Where, v = velocity of sound in air
The apparent frequency of the sound striking the wall is given as:
`v' = (v/(v - v_b))v`
`= (v/(v - 0.03v)) xx 40`
`= 40/0.97` kHz`
This frequency is reflected by the stationary wall (`v_s = 0`) toward the bat.
The frequency (v") of the received sound is given by the relation:
`v" = ((v + v_b)/v) v'`
`= ((v+0.03v)/v) xx 40/0.97`
`= (1.03 xx 40)/ 0.97 = 42.47 kHz`
उत्तर २
Here, the frequency of sound emitted by the bat, υ = 40 kHz. Velocity of bat, υs = 0.03 υ, where υ is velocity of sound. Apparent frequency of sound striking the wall
v' = v/(v - v_s) xx v = v/(v - 0.03v) xx 40 kHz
= `40/0.97 kHz`
This frequency is reflected by the wall and is received by the bat moving towards the wall
So `v_s = 0`
`v_L = 0.03 v`
`v" = (v+v_L)/v xx v'= (v + 0.03v)/v (40/0.97)`
`= 1.03/0.97 xx 40 kHz = 42.47 kHz`
`:. y = (1,1) = 7.5 sin (732.81^@)`
`= 7.5 sin (90 xx 8 + 12.81^@) = 7.5 sin 12.81^@`
= 7.5 xx 0.2217
`= 1.6629 ~~ 1.663 cm`
The velocity of the oscillation at a given point and time is given as:
`v =d/(dt) y (x,t) = d/dt [7.5 sin(0.0050x + 12t + pi/4)]`
`= 7.5 xx 12 cos (0.0050x + 12t + pi/4)`
At x = 1 cm and t = 1 s.
`v = y(1,1)= 90 cos(0.0050x + 12t + pi/4)`
At x = 1 cm and t = 1 s
`v = y(1,1) = 90 cos (12.005 + pi/4)`
= 90 cos(732.81^@) = 90 cos(90xx8+ 12.81)
`= 90 cos(12.81^@)`
`= 90 xx 0.975 = 87.75 "cm/s"`
Now the equation of a propagating wave is given by
`y(x,t) = a sin(kx + wt + phi)`
Where
k = (2pi)/lambda
`:. lambda = (2pi)/k`
And `omega = 2piv`
`:. v = omega/(2pi)`
`Speed, v = vlambda = omega/k`
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