Question

Many natural phenomena produce very high-energy, but inaudible, sound waves at frequencies below 20 Hz (infrasound). During the 2003 eruption of the Fuego volcano in Guatemala, sound waves of frequency 8.0 Hz with a sound level of 120 dBwere recorded. Assume the density of air is 1.2 kg/m3.

What was the maximum displacement of the air molecules produced by the waves?
How much energy would such a wave deliver to a 2.0 m by 3.0 m wall in 1.0 min?

Answer 1

To calculate the maximum displacement of the air molecules and the energy delivered to a wall by infrasound waves produced by a volcano, physics equations relating to wave motion and sound are used. The calculations involve determining the sound intensity from the given decibel level and using this in formulas for displacement amplitude and energy transfer to the wall.

Step-by-step explanation:

Calculating Maximum Displacement and Energy Delivered by Infrasound Waves

Infrasound waves are sound waves with frequencies below the audible range for humans, which is 20 Hz to 20,000 Hz. The scenario described involves infrasound waves generated by a volcanic eruption with a frequency of 8.0 Hz and a sound level of 120 dB, which is inaudible but can still impart energy onto objects. To calculate the maximum displacement of air molecules and the energy delivered to a wall by this wave, we employ physics principles and equations related to wave motion and sound.

Maximum Displacement of Air Molecules

The displacement amplitude of sound waves can be calculated using the formula:

s = √(2 × I / ρ × v)

where s is the maximum displacement, I is the sound intensity, ρ is the density of air, and v is the speed of sound. Intensity (I) can be deduced from the decibel level using I = I0 ×10^(dB/10), with I0 being the reference intensity level of 10⁻¹² W/m². With a sound level of 120 dB, the calculated intensity I substitutes into the displacement formula to find s.

Energy Delivered to a Wall by the Wave

The energy delivered by a wave over some time period can be found using the formula:

E = I × A × t

where E is energy, I is the intensity of the wave, A is the area of the wall, and t is the time in seconds. Given the dimensions of the wall and the duration, we can substitute these values along with the previously calculated sound intensity to find the total energy imparted to the wall.

The exact numerical solutions require specific formula derivations and substitutions not provided in this format, but this explanation gives the framework to compute the needed values.

Answer 2

The energy delivered to the
`\(2.0 \, \text{m} * 3.0 \, \text{m}\)` wall in 1.0 minute by the sound wave is approximately
`\(7.2 * 10^6 \, \text{J}\)`.

To find the maximum displacement of the air molecules produced by the waves, we can use the relationship between sound pressure level (SPL) and maximum particle displacement
`(\(s_{\text{max}}\))` for a given frequency.

The formula relating sound pressure level to maximum displacement for infrasound is given by:

`\[ \text{SPL} = 20 \log_(10)\left(\frac{s_{\text{max}}}{\text{reference pressure}}\right) \]`

Given:

`\(\text{SPL} = 120 \, \text{dB}\)`

Frequency
`\(f = 8.0 \, \text{Hz}\)`

Reference pressure for sound in air
`\(\text{P}_0 = 20 * 10^(-6) \, \text{Pa}\)` (standard reference pressure)

Let's solve for
`\(s_{\text{max}}\)` using the formula:

`\[120 = 20 \log_(10)\left(\frac{s_{\text{max}}}{20 * 10^(-6)}\right)\]`

First, rearrange the equation to solve for
`\(s_{\text{max}}\):`

`\[\frac{s_{\text{max}}}{20 * 10^(-6)} = 10^((120 / 20))\]`

`\[s_{\text{max}} = 20 * 10^(-6) * 10^((120 / 20))\]`

Calculate
`\(s_{\text{max}}\)`:

`\[s_{\text{max}} \approx 0.632 \, \text{m}\]`

The maximum displacement of the air molecules produced by the 8.0 Hz sound wave is approximately
`\(0.632 \, \text{m}\)`.

Next, to find the energy delivered to the wall by this wave, we can use the formula for the energy carried by a sound wave over an area:

`\[ \text{Energy} = \text{Intensity} * \text{Area} * \text{Time} \]`

The intensity (I) of a sound wave is related to the sound pressure level
`(\(\text{SPL}\))`:

`\[ \text{Intensity} = I_0 * 10^{\left(\frac{\text{SPL}}{10}\right)} \]`

Given:

Area (A) of the wall
`= \(2.0 \, \text{m} * 3.0 \, \text{m} = 6.0 \, \text{m}^2\)`

Time (t) = 1.0 minute = 60 seconds

First, let's find the intensity using the sound pressure level:

`\[ \text{Intensity} = I_0 * 10^{\left((120)/(10)\right)} \]`

`\[ \text{Intensity} = I_0 * 10^(12) \]`

Since we're given the density of air, we can use the relationship between intensity, pressure, and density for a sound wave:

`\[ \text{Intensity} = (1)/(2) * \text{pressure}^2 * \text{velocity} \]`

`\[ \text{Pressure} = \sqrt{2 * \text{Intensity} * \text{density}} \]`

Substituting the given values:

`\[ \text{Pressure} = \sqrt{2 * \text{Intensity} * \text{density}} \]`

`\[ \text{Pressure} = \sqrt{2 * (I_0 * 10^(12)) * 1.2 \, \text{kg/m}^3} \]`

Now, we can calculate the pressure and then find the energy delivered to the wall using the formula for energy.

Let's proceed with the calculations.

To find the pressure from the intensity of the sound wave:

`\[\text{Intensity} = I_0 * 10^(12)\]`

Given that
`\(I_0 = 20 * 10^(-6) \, \text{Pa}\)` (standard reference intensity), substitute and calculate the intensity:

`\[\text{Intensity} = 20 * 10^(-6) * 10^(12)\]`

`\[\text{Intensity} = 20000 \, \text{Pa}\]`

Now, calculate the pressure using the intensity and air density:

`\[\text{Pressure} = \sqrt{2 * \text{Intensity} * \text{density}}\]`

`\[\text{Pressure} = √(2 * 20000 * 1.2)\]`

`\[\text{Pressure} = √(48000)\]`

`\[\text{Pressure} \approx 219.09 \, \text{Pa}\]`

Next, calculate the energy delivered to the wall using the formula:

`\[\text{Energy} = \text{Intensity} * \text{Area} * \text{Time}\]`

Given:

Area (A) of the wall
`= \(6.0 \, \text{m}^2\)`

Time
`(\(t\)) = \(60 \, \text{s}\)`

`\[\text{Energy} = 20000 * 6.0 * 60\]`

`\[\text{Energy} = 7200000 \, \text{J}\]`