Question

Assume there is a two-watt source (? = 2 watt) in air and another two-watt source in water. What is the power level ?? in each case? What is the intensity ? at 10 m from the source in each case? What is the sound pressure level ?? at 10 m from the source in each case? What is the source level ?? and transmission loss ? in each case?

Answer 1

Sound intensity is defined to be the power per unit area carried by the sound wave. It is the rate at which energy is transferred by the wave. In equation form,

where P is the power through an area A. The SI unit for I is W/m².

The intensity of a sound wave is related to its amplitude squared by the following relationship:

Here Δp is the pressure variation or pressure amplitude (half the difference between the maximum and minimum pressure in the sound wave) in units of pascals (Pa) or N/m². The kinetic energy of an oscillating element of air due to a traveling sound wave is proportional to its amplitude squared. In this equation, ρ is the density of the material in which the sound wave travels, in units of kg/m³, and v is the speed of sound in the medium, in units of m/s. The pressure variation is proportional to the amplitude of the oscillation, and so I varies as (Δp)² .

The power level is given as:

The source level is given as:

I₀ = 10⁻¹² W/m² is a reference intensity.

For sound source in air:

Power level:

Now calculating the intensity:

So putting this value of I in the second equation:

I considered the air density as 1.29 Kg/m², and velocity of sound in air as 330 m/s.

The source level:

Now the transmission loss is

For sound source in water:

Power level:

Now calculating the intensity:

So putting this value of I in the second equation:

I considered the air density as 1000 Kg/m², and velocity of sound in water at 20^oC as 1481 m/s.

The source level:

Now the transmission loss is

Only the pressure levels differ for differant mediums,

Hope this helps. Please put a like if you liked my answer. Thank you!!