A string of length 0.4 m and mass `10 ^(-2)` kg is tightly clamped at its ends. The tension in the string is 1. 6 N. identical wave pulses are produced at one end at equal intervals of time `Delta t`. The value of `Delta t` which allows construction interference between successive pulses is

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the problem
We have a string of length 0.4 m, mass \(10^{-2}\) kg, and tension 1.6 N. We need to find the time interval \(\Delta t\) that allows for constructive interference between successive wave pulses produced at one end of the string.

### Step 2: Calculate the mass per unit length (\(\mu\))
The mass per unit length \(\mu\) of the string can be calculated using the formula:
\[ \mu = \frac{\text{mass}}{\text{length}} = \frac{10^{-2} \text{ kg}}{0.4 \text{ m}} = 0.025 \text{ kg/m} \]

### Step 3: Calculate the wave speed (\(v\))
The wave speed \(v\) on a string under tension can be calculated using the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \(T\) is the tension in the string. Substituting the values:
\[ v = \sqrt{\frac{1.6 \text{ N}}{0.025 \text{ kg/m}}} = \sqrt{64} = 8 \text{ m/s} \]

### Step 4: Determine the total distance for constructive interference
For constructive interference to occur, the wave must travel to the end of the string and back. Thus, the total distance \(d\) traveled by the wave is:
\[ d = 2 \times \text{length} = 2 \times 0.4 \text{ m} = 0.8 \text{ m} \]

### Step 5: Calculate the time interval \(\Delta t\)
Using the relationship between speed, distance, and time:
\[ v = \frac{d}{\Delta t} \implies \Delta t = \frac{d}{v} \]
Substituting the values we found:
\[ \Delta t = \frac{0.8 \text{ m}}{8 \text{ m/s}} = 0.1 \text{ s} \]

### Final Answer
Thus, the value of \(\Delta t\) that allows for constructive interference between successive pulses is:
\[ \Delta t = 0.1 \text{ seconds} \]

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