A sinusoidal wave travels along a taut string of linear mass density `0.1g//cm`. The particles oscillate along `y-` direction and wave moves in the positive `x-` direction . The amplitude and frequency of oscillation are `2mm` and `50Hz` respectively. The minimum distance between two particles oscillating in the same phase is` 4m`.
The tension in the string is `(` in newton `)`

Options

A. 4000
B. 400
C. 25
D. 250

Explanation

To find the tension in the string, we will follow these steps:

### Step 1: Identify the given data
- Linear mass density (μ) = 0.1 g/cm = 0.1 × 10^-3 kg / 10^-2 m = 0.01 kg/m
- Amplitude (A) = 2 mm = 2 × 10^-3 m
- Frequency (f) = 50 Hz
- Minimum distance between two particles oscillating in the same phase (λ) = 4 m

### Step 2: Calculate the wave velocity (V)
The wave velocity can be calculated using the formula:
\[ V = f \times \lambda \]
Substituting the values:
\[ V = 50 \, \text{Hz} \times 4 \, \text{m} = 200 \, \text{m/s} \]

### Step 3: Relate wave velocity to tension and linear mass density
The wave velocity in a string is also given by the formula:
\[ V = \sqrt{\frac{T}{\mu}} \]
Where:
- T is the tension in the string
- μ is the linear mass density

### Step 4: Rearranging the formula to find tension (T)
Squaring both sides gives:
\[ V^2 = \frac{T}{\mu} \]
Rearranging for T, we get:
\[ T = V^2 \times \mu \]

### Step 5: Substitute the values to find tension
Now substituting the values we have:
\[ T = (200 \, \text{m/s})^2 \times 0.01 \, \text{kg/m} \]
\[ T = 40000 \times 0.01 \]
\[ T = 400 \, \text{N} \]

### Final Answer
The tension in the string is **400 N**.

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