Bat is flying 5m/s while chasing an insect that is flying in the same direction. The bat emits 40kHz chirp and hears a 40.4kHz reflected sound wave. What is the speed of the insect?
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The reflected sound has a higher frequency, so the insect is approaching.
(delta f)/f = 0.4/40 = 0.01 = 2 V/a
The number 2 accounts for Doppler shift incident and reflected sound
a is the sound speed. Assume a = 340 m/s
V = 1.7 m/s
To find the speed of the insect, we can use the Doppler effect equation:
Δf/f = v/c
where Δf is the change in frequency, f is the original frequency of the sound wave, v is the velocity of the source or listener, and c is the speed of sound.
In this scenario, the bat emits a chirp at a frequency of 40kHz and hears a reflected sound wave with a frequency of 40.4kHz.
Let's calculate the change in frequency:
Δf = 40.4kHz - 40kHz
= 0.4kHz
Now, let's plug in the values we have:
Δf/f = v/c
0.4kHz / 40kHz = v / c
Since the speed of sound (c) is approximately 343 m/s, we can rearrange the equation to solve for the velocity of the insect:
v = (0.4kHz / 40kHz) * 343 m/s
= (0.01) * 343 m/s
= 3.43 m/s
Therefore, the speed of the insect is approximately 3.43 m/s.
To solve this problem, we can use the Doppler Effect formula. The Doppler Effect describes the change in frequency of a wave (in this case, sound waves) due to the relative motion between the source of the wave and the listener.
The formula for the observed frequency (f') of a wave, given the emitted frequency (f), the speed of sound (v), and the velocity of the observer (v_obs) relative to the source, is:
f' = (v + v_obs) / (v - v_source) * f
In this case, the bat is the source of the sound waves, and the listener is the bat itself. We know the emitted frequency (f = 40 kHz) and the observed frequency (f' = 40.4 kHz).
Now, let's assign variables to the velocities. Since the bat and the insect are flying in the same direction, their relative velocity is the difference between their speeds. Let's say the speed of the bat is v_bat = 5 m/s. We want to calculate the speed of the insect, so we'll assign v_insect to that velocity.
Using the formula, we can write:
f' = (v + v_obs) / (v - v_source) * f
40.4 kHz = (343 m/s + v_bat) / (343 m/s - v_insect) * 40 kHz
Now, substituting the known values:
40.4 kHz = (343 m/s + 5 m/s) / (343 m/s - v_insect) * 40 kHz
Simplifying the equation:
40.4 kHz = 348 m/s / (343 m/s - v_insect) * 40 kHz
Dividing both sides by 40 kHz:
1.01 = 348 m/s / (343 m/s - v_insect)
Cross-multiplying:
1.01 * (343 m/s - v_insect) = 348 m/s
343 m/s - v_insect = 348 m/s / 1.01
Subtracting 343 m/s from both sides:
-v_insect = 348 m/s / 1.01 - 343 m/s
Multiplying both sides by -1:
v_insect = 343 m/s - 348 m/s / 1.01
Evaluating the expression:
v_insect = 343 m/s - 343.56 m/s
v_insect = -0.56 m/s
Therefore, the speed of the insect is approximately -0.56 m/s. Note that the negative sign indicates that the insect is flying in the opposite direction of the bat.