Question

For a sound coming from a point source, the amplitude of the sound is inversely

proportional to the distance. If the displacement amplitude of an air molecule in a sound
wave is 3.2x10 m at a point 1 m from the source, what would be the displacement
amplitude of the same sound when the distance increases to 4 m?
(Question 3)

Answer 1

The displacement amplitude of the sound at 4 meters is
`8.0* 10^-^1^1` meters as the sound is coming from a point source.

The amplitude of the sound is inversely proportional to the distance, so we can use the formula to relate the amplitudes at different distances:

A1 * d1 = A2 * d2

A1 = the amplitude at the first distance (3.2x10^-10 m)

d1 = the first distance (1 m)

A2 = the amplitude at the second distance (unknown)

d2= s the second distance (4 m)

A1 * d1 = A2 * d2

`3.2* 10^-^1^0` m * 1 m = A2 * 4 m

A2 = (
`3.2* 10^-^1^0` m * 1 m) / 4 m

A2 =
`8.0* 10^-^1^1` m

Answer 2

Answer ) Sound level equation

The intensity of a sound wave is related to its amplitude squared by the following relationship: I=(Δp)22ρvw I = ( Δ p ) 2 2 ρ v w . 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.