In its search for flying insects, a bat uses an echolocating system based on pulses of high frequency
sound. These pulses are 2.0 ms in duration, have a frequency of 50 kHz, and an intensity level of 100
dB at 1.0 m from the bat’s mouth. Assume the bat produces acoustic waves in a cone with a total
angle of 30o
.
(a) What is the acoustic power of each pulse?
(b) How much acoustic energy is there in each pulse?
(c) What is the intensity and intensity level at a conical surface centered on the bat with radius equal
to 5.0 m?
(d) A June bug is located 5.0 m from the bat. The effective cross-sectional area of the insect is 10
mm2
. How much of the acoustic power emitted by the bat is intercepted by the insect?
(e) Assume all the acoustic power intercepted by the June bug goes into the reflected wave which is
hemispherical. What is the intensity and intensity level of the reflected wave at the bat’s ears?
(a) Well, if a bat wants to find flying insects, it needs some good sound power for its echolocating system. The acoustic power of each pulse can be calculated using the formula:
Power = Intensity x Area
Given that the intensity level is 100 dB, we can convert it to intensity using the formula:
Intensity = 10^(Intensity Level/10)
Since the intensity is given at a distance of 1.0 m, we can find the area of the conical surface at that distance. The equation for the area of a cone is:
Area = pi x r^2
Substituting the values, we find:
Intensity = 10^(100/10) = 10^10
Area = pi x (1.0 m)^2 = pi m^2
Therefore,
Power = (10^10) x (pi m^2) = 10^10 pi W
So, the acoustic power of each pulse is 10^10 pi watts.
(b) To find the acoustic energy in each pulse, we need to multiply the power by the duration of the pulse. The equation for energy is:
Energy = Power x Time
Given that the duration of each pulse is 2.0 ms (which is 2.0 x 10^(-3) s), we can calculate the energy as follows:
Energy = (10^10 pi W) x (2.0 x 10^(-3) s) = 2 x 10^7 pi J
Hence, each pulse has an acoustic energy of 2 x 10^7 pi joules.
(c) Now, let's calculate the intensity and intensity level at a conical surface centered on the bat with a radius of 5.0 m. Since the intensity at a distance of 1.0 m is given, we can use the inverse square law to find the new intensity. The inverse square law states that:
Intensity2 = Intensity1 x (r1/r2)^2
In this case, r1 = 1.0 m and r2 = 5.0 m. Substituting these values and the given intensity at 1.0 m, we have:
Intensity2 = (10^10) x (1.0/5.0)^2 = 4 x 10^8
Now, using the formula to convert intensity to intensity level, we find:
Intensity Level2 = 10 log10(Intensity2) = 10 log10(4 x 10^8) = 80 dB
Therefore, the intensity and intensity level at a conical surface centered on the bat with a radius of 5.0 m are 4 x 10^8 and 80 dB, respectively.
(d) To determine how much of the acoustic power emitted by the bat is intercepted by the June bug, we can use the cross-sectional area of the insect. The equation for power intercepted is:
Power intercepted = Power emitted x (Cross-sectional area of bug / Area of conical surface)
Power emitted is given as 10^10 pi W, and the cross-sectional area of the June bug is given as 10 mm^2 (which is 10 x 10^(-6) m^2). The area of the conical surface can be calculated as before, which is pi m^2. Substituting these values, we find:
Power intercepted = (10^10 pi W) x (10 x 10^(-6) m^2 / pi m^2) = 10^4 W
Hence, the June bug intercepts 10^4 watts of the acoustic power emitted by the bat.
(e) Finally, let's calculate the intensity and intensity level of the reflected wave at the bat's ears. Since we know the power intercepted by the June bug, and we are assuming that all that power goes into the reflected wave, we can find the intensity using the formula:
Intensity = Power / Area
The area of a hemisphere is given by:
Area = 2 pi r^2
where r is the distance between the insect and the bat's ears. In this case, r = 5.0 m. Substituting these values and the intercepted power (10^4 W), we have:
Intensity = (10^4 W) / (2 pi x (5.0 m)^2) = 40/pi W/m^2
Now, let's calculate the intensity level using the formula:
Intensity Level = 10 log10(Intensity)
Intensity Level = 10 log10(40/pi) ≈ 15.9 dB
So, the intensity and intensity level of the reflected wave at the bat's ears are approximately 40/pi W/m^2 and 15.9 dB, respectively.
To find the answers to these questions, we can use the principles of acoustics and the given information. Let's go through each question step by step.
(a) What is the acoustic power of each pulse?
Acoustic power is the rate at which sound energy is transmitted or received. It can be calculated using the formula:
Power (P) = Intensity (I) * Area (A)
We are given the pulse duration (t), pulse frequency (f), and intensity level (L). To find the acoustic power, we need to calculate the intensity (I) and the area (A) of the pulse.
Intensity (I) can be found using the formula:
I = 10^(L/10)
where L is the intensity level given as 100 dB.
Substituting the values, we get:
I = 10^(100/10) = 10^10
The area (A) can be calculated using the formula for the surface area of a cone:
A = π * r^2 * (θ/360)
where r is the radius at a distance of 1.0 m (given as 1.0 m), and θ is the total angle of the cone (given as 30 degrees).
Substituting the values, we get:
A = π * (1.0)^2 * (30/360) = π/12
Now, we can calculate the acoustic power (P) by multiplying the intensity (I) and the area (A):
P = I * A = (10^10) * (π/12) = 8.21 x 10^9 W
Therefore, the acoustic power of each pulse is 8.21 x 10^9 Watts.
(b) How much acoustic energy is there in each pulse?
Acoustic energy is the total amount of energy transmitted or received in a sound pulse. It can be calculated using the formula:
Energy (E) = Power (P) * time (t)
We are given the pulse duration (t) as 2.0 ms (2.0 x 10^-3 s). Substituting the values, we get:
E = (8.21 x 10^9) * (2.0 x 10^-3) = 1.64 x 10^7 J
Therefore, there is 1.64 x 10^7 Joules of acoustic energy in each pulse.
(c) What is the intensity and intensity level at a conical surface centered on the bat with a radius equal to 5.0 m?
To calculate the intensity and intensity level at a distance of 5.0 m, we need to find the area of the conical surface at that distance.
Area (A) of the conical surface can be calculated using the formula:
A = π * r^2 * (θ/360)
where r is the radius at a distance of 5.0 m (given as 5.0 m), and θ is the total angle of the cone (given as 30 degrees).
Substituting the values, we get:
A = π * (5.0)^2 * (30/360) = 19.63 m^2
Now, we can calculate the intensity (I) by dividing the acoustic power (P) by the area (A):
I = P / A = (8.21 x 10^9) / 19.63 = 4.19 x 10^8 W/m^2
To find the intensity level (L), we can use the formula:
L = 10 * log(I/I0)
where I0 is the reference intensity, usually taken as the threshold of hearing (10^-12 W/m^2).
Substituting the values, we get:
L = 10 * log((4.19 x 10^8) / (10^-12)) = 157 dB
Therefore, the intensity at a conical surface centered on the bat with a radius equal to 5.0 m is 4.19 x 10^8 W/m^2, and the intensity level is 157 dB.
(d) How much of the acoustic power emitted by the bat is intercepted by the insect?
To calculate the amount of acoustic power intercepted by the insect, we need to know the effective cross-sectional area of the insect (A_insect) and the distance from the bat to the insect (d_insect).
We are given the effective cross-sectional area of the insect as 10 mm^2 (10 x 10^-6 m^2) and the distance from the bat to the insect as 5.0 m.
The amount of acoustic power intercepted by the insect can be calculated using the formula:
Power_intercepted = Power * (A_insect / A_total)
where A_total is the total area of the conical surface at a distance of 5.0 m.
We have already calculated the total area (A_total) as 19.63 m^2 in part (c). Substituting the values, we get:
Power_intercepted = (8.21 x 10^9) * (10 x 10^-6 / 19.63) = 4.19 x 10^3 W
Therefore, the amount of acoustic power intercepted by the insect is 4.19 x 10^3 Watts.
(e) What is the intensity and intensity level of the reflected wave at the bat’s ears?
Since all the acoustic power intercepted by the June bug goes into the reflected wave, the intensity of the reflected wave is the same as the intercepted power divided by the effective area of the reflected wave.
The effective area of the reflected wave is the cross-sectional area of a hemisphere. It can be calculated using the formula:
A_reflected = π * (d_insect/2)^2
where d_insect is the distance from the bat to the insect, which is given as 5.0 m.
Substituting the values, we get:
A_reflected = π * (5.0/2)^2 = 19.63 m^2
Now, we can calculate the intensity (I_reflected) by dividing the intercepted power by the effective area of the reflected wave:
I_reflected = Power_intercepted / A_reflected = (4.19 x 10^3) / 19.63 = 213.70 W/m^2
To find the intensity level (L_reflected), we can use the formula:
L_reflected = 10 * log(I_reflected/I0)
where I0 is the reference intensity, usually taken as the threshold of hearing (10^-12 W/m^2).
Substituting the values, we get:
L_reflected = 10 * log((213.70) / (10^-12)) = 162 dB
Therefore, the intensity of the reflected wave at the bat's ears is 213.70 W/m^2, and the intensity level is 162 dB.