A uniform string of length 10 m is suspended from a rigid support A short wave pulse is introduced at its lowest end it starts moving up the string the time taken to reach the support is (take `g=10m//sec^(2)`)

Options

A. 1 sec
B. 2 sec
C. `2sqrt(2)`
D. 4 sec

Explanation

To solve the problem of how long it takes for a wave pulse to travel up a uniform string of length 10 m, we can follow these steps:

### Step 1: Understand the Wave Velocity on the String
The velocity \( v \) of a wave on a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the linear mass density of the string.

### Step 2: Determine the Tension in the String
The tension \( T \) in the string at a distance \( x \) from the bottom can be expressed as:
\[ T = m g = \mu x g \]
where \( m = \mu x \) is the mass of the string below the point where the wave is introduced, and \( g \) is the acceleration due to gravity.

### Step 3: Substitute Tension into the Wave Velocity Formula
Substituting the expression for tension into the velocity formula gives:
\[ v = \sqrt{\frac{\mu x g}{\mu}} = \sqrt{g x} \]

### Step 4: Relate Velocity and Acceleration
The acceleration \( a \) of the wave pulse can be expressed using the chain rule:
\[ a = v \frac{dv}{dx} \]
Substituting \( v = \sqrt{g x} \) into this expression gives:
\[ a = \sqrt{g x} \cdot \frac{1}{2\sqrt{g x}} \cdot g = \frac{g}{2} \]

### Step 5: Use the Kinematic Equation to Find Time
We can use the kinematic equation:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( s \) is the distance traveled (which is the length of the string \( L = 10 \, m \)), \( u = 0 \) (initial velocity), and \( a = \frac{g}{2} \).

Substituting these values into the equation gives:
\[ 10 = 0 \cdot t + \frac{1}{2} \cdot \frac{g}{2} \cdot t^2 \]
This simplifies to:
\[ 10 = \frac{g}{4} t^2 \]

### Step 6: Solve for Time \( t \)
Rearranging the equation gives:
\[ t^2 = \frac{40}{g} \]
Substituting \( g = 10 \, m/s^2 \):
\[ t^2 = \frac{40}{10} = 4 \]
Taking the square root:
\[ t = \sqrt{4} = 2 \, seconds \]

### Final Answer
The time taken for the wave pulse to reach the support is \( 2 \, seconds \).

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