A string of length 0.4 m and mass `10^(-2)` kg is clamped at its ends. The tension in the string is 1.6 N. When a pulse travels along the string the shape of the string is found to be the same at times t and `t+trianglet`. The value `trianglet` is

Options

A. 0.05s
B. 0.1s
C. 0.2s
D. 0.4s

Answer

To solve the problem, we need to determine the value of \( \Delta t \) when the shape of the string is the same at times \( t \) and \( t + \Delta t \). This implies that \( \Delta t \) corresponds to the time period of the wave traveling along the string.

### Step-by-Step Solution:

1. **Identify the parameters given in the problem:**
- Length of the string, \( L = 0.4 \, \text{m} \)
- Mass of the string, \( m = 10^{-2} \, \text{kg} \)
- Tension in the string, \( T = 1.6 \, \text{N} \)

2. **Calculate the linear mass density (\( \mu \)) of the string:**
\[ \mu = \frac{m}{L} = \frac{10^{-2} \, \text{kg}}{0.4 \, \text{m}} = 0.025 \, \text{kg/m} \]

3. **Calculate the wave speed (\( v \)) in the string using the formula:**
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the values:
\[ v = \sqrt{\frac{1.6 \, \text{N}}{0.025 \, \text{kg/m}}} = \sqrt{64} = 8 \, \text{m/s} \]

4. **Determine the wavelength (\( \lambda \)) of the wave:**
Since the string is clamped at both ends, the fundamental frequency corresponds to a wavelength equal to twice the length of the string:
\[ \lambda = 2L = 2 \times 0.4 \, \text{m} = 0.8 \, \text{m} \]

5. **Calculate the time period (\( T \)) of the wave:**
The time period is given by the formula:
\[ T = \frac{\lambda}{v} \]
Substituting the values:
\[ T = \frac{0.8 \, \text{m}}{8 \, \text{m/s}} = 0.1 \, \text{s} \]

6. **Determine \( \Delta t \):**
Since the shape of the string is the same at times \( t \) and \( t + \Delta t \), we have:
\[ \Delta t = T = 0.1 \, \text{s} \]

### Final Answer:
\[ \Delta t = 0.1 \, \text{s} \]