What is the frequency of the fundamental note emitted by a wire of length 2 m and radius 1 mm under a tension of 200 N. Density of the wire is `"1600 kg/m"^3.`

Answer

To find the frequency of the fundamental note emitted by the wire, we can follow these steps:

### Step 1: Identify the given values
- Length of the wire (L) = 2 m
- Radius of the wire (r) = 1 mm = \(1 \times 10^{-3}\) m
- Tension in the wire (T) = 200 N
- Density of the wire (ρ) = 1600 kg/m³

### Step 2: Calculate the cross-sectional area (A) of the wire
The cross-sectional area of a wire with a circular cross-section is given by the formula:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (1 \times 10^{-3})^2 = \pi \times 10^{-6} \text{ m}^2 \]

### Step 3: Calculate the mass per unit length (μ)
The mass per unit length (μ) can be calculated using the formula:
\[ \mu = \frac{\text{mass}}{\text{length}} = \frac{\text{density} \times \text{volume}}{\text{length}} \]
Since volume \(V = A \times L\), we can express μ as:
\[ \mu = \rho \times A \]
Substituting the values:
\[ \mu = 1600 \times \pi \times 10^{-6} \text{ kg/m} \]

### Step 4: Substitute μ into the frequency formula
The frequency of the fundamental note (f) is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Substituting the values we have:
\[ f = \frac{1}{2 \times 2} \sqrt{\frac{200}{1600 \times \pi \times 10^{-6}}} \]

### Step 5: Simplify the expression
Calculating the denominator:
\[ \mu = 1600 \times \pi \times 10^{-6} \approx 5.0265 \times 10^{-3} \text{ kg/m} \]
Now substituting this back into the frequency formula:
\[ f = \frac{1}{4} \sqrt{\frac{200}{5.0265 \times 10^{-3}}} \]

### Step 6: Calculate the square root
Calculating the value inside the square root:
\[ \frac{200}{5.0265 \times 10^{-3}} \approx 39763.9 \]
Taking the square root:
\[ \sqrt{39763.9} \approx 199.4 \]

### Step 7: Calculate the frequency
Now substituting back to find f:
\[ f = \frac{1}{4} \times 199.4 \approx 49.85 \text{ Hz} \]
Rounding it off gives:
\[ f \approx 50 \text{ Hz} \]

### Final Answer
The frequency of the fundamental note emitted by the wire is approximately **50 Hz**.

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