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# Solution for 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? - CBSE (Science) Class 11 - Physics

ConceptReflection of Waves Standing Waves and Normal Modes

#### Question

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?

#### Solution 1

Ultrasonic beep frequency emitted by the bat, ν = 40 kHz

Velocity of the bat, vb = 0.03 v

Where, = 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

#### Solution 2

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`

Is there an error in this question or solution?

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Solution for question: 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? concept: Reflection of Waves - Standing Waves and Normal Modes. For the course CBSE (Science)
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