At t=0,a transverse wave pulse travelling in the positive x direction with a speed of `2 m//s` in a wire is described by the function `y=6//x^(2)` given that `x!=0`. Transverse velocity of a particle at x=2 m and t= 2 s is

Options

A. `3 m//s`
B. `-3 m//s`
C. `8 m//s`
D. `-8 m//s`

Explanation

To solve the problem, we need to find the transverse velocity of a particle at \( x = 2 \, \text{m} \) and \( t = 2 \, \text{s} \) for the given wave function.

### Step 1: Write the wave function at time \( t \)

Given the wave function at \( t = 0 \):
\[ y = \frac{6}{x^2} \]

Since the wave is moving in the positive x-direction with a speed of \( 2 \, \text{m/s} \), we can modify the wave function to account for time. The general form of the wave function for a wave traveling in the positive x-direction is:
\[ y(x, t) = \frac{6}{(x - vt)^2} \]
where \( v \) is the speed of the wave. Substituting \( v = 2 \, \text{m/s} \):
\[ y(x, t) = \frac{6}{(x - 2t)^2} \]

### Step 2: Find the transverse velocity

The transverse velocity \( v_t \) of a particle in the wave is given by:
\[ v_t = \frac{\partial y}{\partial t} \]

To find this, we first need to differentiate \( y(x, t) \) with respect to \( t \):
\[ y(x, t) = \frac{6}{(x - 2t)^2} \]

Using the chain rule:
\[ \frac{\partial y}{\partial t} = 6 \cdot \frac{d}{dt} \left( (x - 2t)^{-2} \right) \]
\[ = 6 \cdot (-2) \cdot (x - 2t)^{-3} \cdot \frac{d}{dt}(x - 2t) \]
\[ = 6 \cdot (-2) \cdot (x - 2t)^{-3} \cdot (-2) = \frac{24}{(x - 2t)^3} \]

### Step 3: Substitute \( x = 2 \, \text{m} \) and \( t = 2 \, \text{s} \)

Now, we need to evaluate \( v_t \) at \( x = 2 \, \text{m} \) and \( t = 2 \, \text{s} \):
\[ v_t = \frac{24}{(2 - 2 \cdot 2)^3} = \frac{24}{(2 - 4)^3} = \frac{24}{(-2)^3} = \frac{24}{-8} = -3 \, \text{m/s} \]

### Final Answer

The transverse velocity of a particle at \( x = 2 \, \text{m} \) and \( t = 2 \, \text{s} \) is:
\[ \boxed{-3 \, \text{m/s}} \]