The linear density of a vibrating string is `1.3 xx 10^(-4) kg//m` A transverse wave is propagating on the string and is described by the equation `y= 0.021 sin (x + 30 t)` where x and y are measured in meter and t`t` in second the tension in the string is :-

Options

A. 0.12 N
B. 0.48 N
C. 1.20 N
D. 4.80 N

Answer

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

### Step 1: Identify the given values
- Linear density (μ) = \(1.3 \times 10^{-4} \, \text{kg/m}\)
- Wave equation: \(y = 0.021 \sin(x + 30t)\)

### Step 2: Extract the angular frequency (ω) and wave number (k)
From the wave equation \(y = A \sin(kx + \omega t)\):
- The angular frequency (ω) is the coefficient of \(t\), which is \(30 \, \text{rad/s}\).
- The wave number (k) is the coefficient of \(x\). Since there is no explicit coefficient, we can assume \(k = 1 \, \text{rad/m}\).

### Step 3: Calculate the tension (T) using the formula
The tension in the string can be calculated using the formula:
\[ T = \mu \cdot \omega^2 \cdot k^2 \]

### Step 4: Substitute the values into the formula
Substituting the values we found:
- \(\mu = 1.3 \times 10^{-4} \, \text{kg/m}\)
- \(\omega = 30 \, \text{rad/s}\)
- \(k = 1 \, \text{rad/m}\)

Now, we calculate:
\[ T = (1.3 \times 10^{-4}) \cdot (30^2) \cdot (1^2) \]

### Step 5: Perform the calculations
Calculating \(30^2\):
\[ 30^2 = 900 \]
Now substituting back:
\[ T = (1.3 \times 10^{-4}) \cdot 900 \]
\[ T = 1.17 \times 10^{-1} \, \text{N} = 0.117 \, \text{N} \]

### Step 6: Round the answer
Rounding \(0.117 \, \text{N}\) gives approximately \(0.12 \, \text{N}\).

### Final Answer
The tension in the string is approximately \(0.12 \, \text{N}\).

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