A string of mass `0.250` kg is under a tension of 2 N. The length of the stretched string is 2 m. A transverse wave disturbance starts at one end of the string . How long will the disturbance take to reach the other end ?

Options

A. `1,5` s
B. `1. 5` s
C. 1 s
D. `0.5` s

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Mass of the string (m) = 0.250 kg
- Tension in the string (T) = 2 N
- Length of the string (L) = 2 m

### Step 2: Calculate the linear density (μ)
The linear density (μ) is defined as the mass per unit length of the string. It can be calculated using the formula:
\[ \mu = \frac{m}{L} \]
Substituting the values:
\[ \mu = \frac{0.250 \, \text{kg}}{2 \, \text{m}} = 0.125 \, \text{kg/m} \]

### Step 3: Calculate the velocity (v) of the wave
The velocity of a transverse wave on a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the values of tension and linear density:
\[ v = \sqrt{\frac{2 \, \text{N}}{0.125 \, \text{kg/m}}} \]
Calculating the value:
\[ v = \sqrt{16} = 4 \, \text{m/s} \]

### Step 4: Calculate the time (t) taken for the disturbance to travel the length of the string
The time taken for the wave to travel a distance is given by the formula:
\[ t = \frac{L}{v} \]
Substituting the values:
\[ t = \frac{2 \, \text{m}}{4 \, \text{m/s}} = 0.5 \, \text{s} \]

### Final Answer
The disturbance will take **0.5 seconds** to reach the other end of the string.

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