A point `A` is located at a distance `r = 1.5 m` from a point source of sound of frequency `600 H_(Z)`. The power of the source is `0.8 W`. Speed of sound in air is `340 m//s` and density of air is `1.29 kg//m^(3)`. Find at the point `A`, (a) the pressure oscillation amplitude`(Deltap)_(m)` (b) the displacement oscillation amplitude `A`.

Explanation

To solve the problem step by step, we will find both the pressure oscillation amplitude (ΔP_m) and the displacement oscillation amplitude (A) at point A, which is located at a distance \( r = 1.5 \, m \) from a point source of sound.

### Given Data:
- Frequency, \( f = 600 \, Hz \)
- Distance, \( r = 1.5 \, m \)
- Power of the source, \( P = 0.8 \, W \)
- Speed of sound in air, \( v = 340 \, m/s \)
- Density of air, \( \rho = 1.29 \, kg/m^3 \)

### Step 1: Calculate the Intensity of the Sound
The intensity \( I \) of the sound at a distance \( r \) from a point source is given by:

\[ I = \frac{P}{4\pi r^2} \]

Substituting the values:

\[ I = \frac{0.8}{4 \times \pi \times (1.5)^2} \]

Calculating \( I \):

\[ I = \frac{0.8}{4 \times 3.14 \times 2.25} \approx \frac{0.8}{28.26} \approx 0.0283 \, W/m^2 \]

### Step 2: Calculate the Displacement Oscillation Amplitude \( A \)
The intensity can also be expressed in terms of the displacement amplitude \( A \):

\[ I = \frac{1}{2} \rho v \omega^2 A^2 \]

Where \( \omega = 2\pi f \). First, we calculate \( \omega \):

\[ \omega = 2\pi \times 600 \approx 3768.6 \, rad/s \]

Now substituting \( I \) into the intensity formula:

\[ 0.0283 = \frac{1}{2} \times 1.29 \times 340 \times (3768.6)^2 \times A^2 \]

Rearranging to solve for \( A^2 \):

\[ A^2 = \frac{2 \times 0.0283}{1.29 \times 340 \times (3768.6)^2} \]

Calculating \( A^2 \):

\[ A^2 \approx \frac{0.0566}{1.29 \times 340 \times 142,157,000} \approx \frac{0.0566}{61,000,000} \approx 9.27 \times 10^{-9} \]

Taking the square root to find \( A \):

\[ A \approx \sqrt{9.27 \times 10^{-9}} \approx 3.05 \times 10^{-5} \, m \]

### Step 3: Calculate the Pressure Oscillation Amplitude \( \Delta P_m \)
The pressure oscillation amplitude \( \Delta P_m \) is given by:

\[ \Delta P_m = \rho v \omega A \]

Substituting the values:

\[ \Delta P_m = 1.29 \times 340 \times 3768.6 \times (3.05 \times 10^{-5}) \]

Calculating \( \Delta P_m \):

\[ \Delta P_m \approx 1.29 \times 340 \times 3768.6 \times 3.05 \times 10^{-5} \approx 4.98 \times 10^{-6} \, N/m^2 \]

### Final Answers:
(a) The pressure oscillation amplitude \( \Delta P_m \approx 4.98 \times 10^{-6} \, N/m^2 \)

(b) The displacement oscillation amplitude \( A \approx 3.05 \times 10^{-5} \, m \)

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