To solve the problem of finding the time taken by a wave pulse to travel from the lighter end to the heavier end of a non-uniform wire, we can follow these steps:
### Step 1: Understand the given parameters
We have:
- Length of the wire, \( L = 10 \, \text{m} \)
- Mass per unit length, \( \mu = \mu_0 + ax \)
- Tension in the wire, \( T = 100 \, \text{N} \)
- Given values: \( \mu_0 = 10^{-2} \, \text{kg/m} \) and \( a = 9 \times 10^{-3} \, \text{kg/m}^2 \)
### Step 2: Write the expression for wave velocity
The velocity of a wave pulse in a string is given by:
\[
v = \sqrt{\frac{T}{\mu}}
\]
Since \( \mu \) varies with \( x \), we can express the velocity as:
\[
v(x) = \sqrt{\frac{T}{\mu_0 + ax}}
\]
### Step 3: Set up the equation for time
The time \( dt \) taken to travel a small distance \( dx \) is given by:
\[
dt = \frac{dx}{v(x)} = \frac{dx}{\sqrt{\frac{T}{\mu_0 + ax}}}
\]
This can be rewritten as:
\[
dt = \sqrt{\frac{\mu_0 + ax}{T}} \, dx
\]
### Step 4: Integrate to find total time
To find the total time \( t \) taken to travel from \( x = 0 \) to \( x = 10 \):
\[
t = \int_0^{10} \sqrt{\frac{\mu_0 + ax}{T}} \, dx
\]
Substituting the values of \( T \), \( \mu_0 \), and \( a \):
\[
t = \int_0^{10} \sqrt{\frac{\mu_0 + ax}{100}} \, dx
\]
### Step 5: Calculate the integral
Substituting \( \mu_0 = 10^{-2} \) and \( a = 9 \times 10^{-3} \):
\[
t = \int_0^{10} \sqrt{\frac{10^{-2} + 9 \times 10^{-3} x}{100}} \, dx
\]
\[
= \frac{1}{10} \int_0^{10} \sqrt{10^{-2} + 9 \times 10^{-3} x} \, dx
\]
### Step 6: Evaluate the integral
Let \( u = 10^{-2} + 9 \times 10^{-3} x \) then \( du = 9 \times 10^{-3} dx \) or \( dx = \frac{du}{9 \times 10^{-3}} \).
When \( x = 0 \), \( u = 10^{-2} \) and when \( x = 10 \), \( u = 10^{-2} + 9 \times 10^{-2} = 10^{-1} \).
Now, the integral becomes:
\[
t = \frac{1}{10} \int_{10^{-2}}^{10^{-1}} \sqrt{u} \cdot \frac{du}{9 \times 10^{-3}} = \frac{1}{90 \times 10^{-3}} \int_{10^{-2}}^{10^{-1}} u^{1/2} \, du
\]
Evaluating the integral:
\[
\int u^{1/2} \, du = \frac{2}{3} u^{3/2}
\]
Thus,
\[
t = \frac{1}{90 \times 10^{-3}} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{10^{-2}}^{10^{-1}} = \frac{1}{90 \times 10^{-3}} \cdot \frac{2}{3} \left( (10^{-1})^{3/2} - (10^{-2})^{3/2} \right)
\]
Calculating the values:
\[
= \frac{1}{90 \times 10^{-3}} \cdot \frac{2}{3} \left( 10^{-3/2} - 10^{-3} \right)
\]
\[
= \frac{1}{90 \times 10^{-3}} \cdot \frac{2}{3} \left( 10^{-1.5} - 10^{-3} \right)
\]
\[
= \frac{2}{270 \times 10^{-3}} \left( 10^{-1.5} - 10^{-3} \right)
\]
### Step 7: Final calculation
Calculating the numerical values gives:
\[
t \approx 0.227 \, \text{s}
\]
### Final Answer
The time taken by the pulse to travel from the lighter end to the heavier end of the wire is approximately **0.227 seconds**.
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