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 30degree.
(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) To find the acoustic power of each pulse, we can use the formula:

Power = Intensity × Area

The area of the wavefront can be calculated using the formula for a cone:

Area = π × r^2 × sin(θ/2)

Where r is the radius of the cone's base and θ is the total angle of the cone.

Given that the pulse has a duration of 2.0 ms, we can calculate the period as follows:

T = 2.0 ms = 2.0 × 10^(-3) s

The intensity of the pulse at 1.0 m from the bat's mouth is 100 dB, which can be converted to Watts/m^2 using the formula:

Intensity (W/m^2) = 10^(Intensity(dB)/10)

Therefore:

Intensity (W/m^2) = 10^(100/10) = 10^10 W/m^2

Substituting the given values into the area formula:

Area = π × (1.0)^2 × sin(30/2) = π × sin(15)

Now, we can calculate the power:

Power = Intensity × Area = (10^10) × (π × sin(15))

(b) To find the acoustic energy in each pulse, we need to multiply the power by the time:

Energy = Power × Time = Power × Period

Substituting the values we calculated:

Energy = (10^10) × (π × sin(15)) × (2.0 × 10^(-3))

(c) At a conical surface centered on the bat with a radius of 5.0 m, we need to calculate the intensity and intensity level.

The intensity is given by:

Intensity = Power / Area

Area = π × r^2 × sin(θ/2), where r = 5.0 m and θ = 30 degrees.

Intensity (W/m^2) = Power / (π × (5.0)^2 × sin(30/2))

To calculate the intensity level, we can use the formula:

Intensity level (dB) = 10 × log10(Intensity / (10^(-12) W/m^2))

Substituting the values we calculated:

Intensity level = 10 × log10(Intensity / (10^(-12)))

(d) To find how much acoustic power emitted by the bat is intercepted by the June bug, we need to calculate the power intercepted.

The power intercepted is given by:

Power intercepted = Intensity × Area (cross-sectional area of the insect)

Given that the effective cross-sectional area of the insect is 10 mm^2, we need to convert it to m^2:

Area (m^2) = 10 × 10^(-6) m^2

Now, we can calculate the power intercepted:

Power intercepted = Intensity × Area (m^2)

(e) To find the intensity and intensity level of the reflected wave at the bat's ears, we need to consider that the power intercepted by the June bug goes into the reflected wave, which is hemispherical.

The intensity of the reflected wave is given by:

Intensity (W/m^2) = Power intercepted / (2 × π × r^2), where r is the distance from the bat to the June bug.

To calculate the intensity level, we can use the formula:

Intensity level (dB) = 10 × log10(Intensity / (10^(-12) W/m^2))

(a) To calculate the acoustic power of each pulse, we need to use the formula for power:

Power = Intensity * Area

First, we need to convert the sound intensity from decibels (dB) to watts per square meter (W/m^2). The formula for this conversion is:

Intensity (W/m^2) = 10^(dB/10)

We can calculate the intensity using the given intensity level of 100 dB:

Intensity = 10^(100/10) = 10^10 W/m^2

Next, we need to calculate the area of the cone at a distance of 1.0 m from the bat's mouth. The formula for the surface area of a cone is:

Area = π * r * (r + L)

Given that the cone has a total angle of 30 degrees and the radius (r) is 1.0 m, we can calculate L (the slant height) using trigonometry.

L = r / tan(angle/2) = 1.0 / tan(30/2) = 1.155 m

Now we can calculate the surface area:

Area = π * 1.0 * (1.0 + 1.155) = 6.267 m^2

Finally, we can calculate the acoustic power:

Power = Intensity * Area = 10^10 W/m^2 * 6.267 m^2 = 6.267 × 10^10 W

Therefore, the acoustic power of each pulse is 6.267 × 10^10 watts.

(b) To calculate the acoustic energy of each pulse, we need to use the formula:

Energy = Power * Time

Given that the pulse duration is 2.0 ms (0.002 s), we can calculate the energy:

Energy = 6.267 × 10^10 W * 0.002 s = 1.253 × 10^8 J

Therefore, the acoustic energy of each pulse is 1.253 × 10^8 joules.

(c) To calculate the intensity and intensity level at a conical surface centered on the bat with a radius of 5.0 m, we need to use the inverse square law for sound:

Intensity (r) = Intensity (1.0 m) * (1.0 m / r)^2

Intensity (5.0 m) = 10^10 W/m^2 * (1.0 m / 5.0 m)^2 = 4 × 10^8 W/m^2

To calculate the intensity level, we use the formula:

Intensity Level (dB) = 10 * log10(Intensity / Iref)

Where Iref is the reference intensity of 10^(-12) W/m^2. Substituting the values, we get:

Intensity Level (5.0 m) = 10 * log10(4 × 10^8 / 10^(-12)) = 150 dB

Therefore, the intensity at the conical surface centered on the bat with a radius of 5.0 m is 4 × 10^8 W/m^2 and the intensity level is 150 dB.

(d) To calculate the acoustic power intercepted by the insect, we need to use the formula:

Power intercepted = Intensity at insect location * Area of insect

Given that the insect is located 5.0 m from the bat and has a cross-sectional area of 10 mm^2 (0.00001 m^2), we can calculate the power intercepted:

Power intercepted = 10^10 W/m^2 * 0.00001 m^2 = 100 W

Therefore, the insect intercepts 100 watts of the acoustic power emitted by the bat.

(e) If all the acoustic power intercepted by the June bug goes into the reflected wave, we can calculate the intensity and intensity level of the reflected wave at the bat's ears.

Since the reflected wave is hemispherical, the intensity is spread over the surface area of a hemisphere. The surface area of a hemisphere is:

Area = 2πr^2

Given that the bug is located 5.0 m away from the bat, the radius (r) would also be 5.0 m. We can calculate the area:

Area = 2π * 5.0^2 = 157.08 m^2

To calculate the intensity, we divide the power intercepted by the area:

Intensity = Power intercepted / Area = 100 W / 157.08 m^2 = 0.637 W/m^2

To convert the intensity to decibels, we use the formula:

Intensity Level (dB) = 10 * log10(Intensity / Iref)

Substituting the values, we get:

Intensity Level (reflected wave) = 10 * log10(0.637 / 10^(-12)) = 71 dB

Therefore, the intensity of the reflected wave at the bat's ears is 0.637 W/m^2 and the intensity level is 71 dB.