(a) A bat locates insects by emitting ultrasonic “chirps” and then listening for echoes from
the bugs. Suppose a bat chirp has a frequency of 25 kHz. How fast would the bat have to
fly, and in what direction, for you to just barely be able to hear the chirp at 20 kHz?
(b) If this sound has an intensity of 2.0 × 10−3 W/m2
. Your eardrum is 6.0 mm in diameter.
How much energy will be transferred to your eardrum while listening to the bat for 1.0 min
(a) The frequency of a sound wave is affected by the motion of the source relative to the observer, known as the Doppler effect. The formula to calculate the observed frequency is given by
f' = f(v + vo) / (v - vs)
where
f' is the observed frequency,
f is the actual frequency (25 kHz),
v is the speed of sound in air (assumed to be approximately 343 m/s),
vo is the velocity of the observer (0 m/s since you are stationary),
vs is the velocity of the source (unknown).
Rearranging the formula to solve for vs, we get
vs = (f'v - fv) / (f' + f)
Plugging in the observed frequency f' of 20 kHz and the other known values, we have
vs = (20,000 Hz * 343 m/s - 25,000 Hz * 343 m/s) / (20,000 Hz + 25,000 Hz)
Simplifying this expression gives us
vs ≈ -86 m/s
Therefore, the bat would have to fly towards you at a speed of approximately 86 m/s.
(b) The energy (E) transferred to your eardrum can be calculated by multiplying the intensity (I) by the area (A) and the time (t):
E = I * A * t
First, we need to calculate the area of your eardrum. The diameter of your eardrum is 6.0 mm, which corresponds to a radius of 3.0 mm or 0.003 m. The area is given by:
A = π * r^2
A = π * (0.003 m)^2
Calculating A gives us:
A ≈ 2.83 x 10^-5 m^2
Next, we can calculate the energy transferred using the given intensity, area, and time:
E = (2.0 x 10^-3 W/m^2) * (2.83 x 10^-5 m^2) * (60 s)
Calculating E gives us:
E ≈ 2.28 x 10^-5 J
Therefore, approximately 2.28 x 10^-5 joules of energy would be transferred to your eardrum while listening to the bat for 1.0 minute.