The amplitude of the sound wave emitted by a source is `10^(-3)m.` If the frequency of sound is 600 Hz, what is (i) the energy per unit volume and (ii) the intensity of sound ? Speed of sound = 340 m/s, Density of air `="1.29 kg/m"^3`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Amplitude (A) = \(10^{-3} \, \text{m}\)
- Frequency (f) = \(600 \, \text{Hz}\)
- Speed of sound (v) = \(340 \, \text{m/s}\)
- Density of air (\(\rho\)) = \(1.29 \, \text{kg/m}^3\)

### Step 2: Calculate angular frequency (\(\omega\))
The angular frequency is given by the formula:
\[ \omega = 2\pi f \]
Substituting the frequency:
\[ \omega = 2 \times \pi \times 600 \approx 3769.91 \, \text{rad/s} \]

### Step 3: Calculate energy per unit volume (sound density)
The energy per unit volume (E) is given by the formula:
\[ E = \frac{1}{2} \rho A^2 \omega^2 \]
Substituting the known values:
\[ E = \frac{1}{2} \times 1.29 \, \text{kg/m}^3 \times (10^{-3} \, \text{m})^2 \times (3769.91 \, \text{rad/s})^2 \]
Calculating:
\[ E = \frac{1}{2} \times 1.29 \times 10^{-6} \times 142,00000 \approx 9.1575 \, \text{J/m}^3 \]

### Step 4: Calculate intensity of sound
The intensity (I) of sound is given by the formula:
\[ I = \frac{1}{2} \rho v A^2 \omega^2 \]
Substituting the known values:
\[ I = \frac{1}{2} \times 1.29 \, \text{kg/m}^3 \times 340 \, \text{m/s} \times (10^{-3} \, \text{m})^2 \times (3769.91 \, \text{rad/s})^2 \]
Calculating:
\[ I = \frac{1}{2} \times 1.29 \times 340 \times 10^{-6} \times 142,00000 \approx 3116.74 \, \text{W/m}^2 \]

### Final Answers
1. Energy per unit volume = \(9.1575 \, \text{J/m}^3\)
2. Intensity of sound = \(3116.74 \, \text{W/m}^2\)

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