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 step by step, we will follow these instructions:

### Step 1: Understand the wave function
The wave function at time \( t = 0 \) is given by:
\[ y = \frac{6}{x^2} \]
This represents a transverse wave pulse traveling in the positive x-direction.

### Step 2: Write the general wave equation
Since the wave is traveling in the positive x-direction, we can express the wave function at any time \( t \) as:
\[ y = \frac{6}{(x - vt)^2} \]
where \( v \) is the speed of the wave. Given that the speed \( v = 2 \, \text{m/s} \), we can substitute this into the equation:
\[ y = \frac{6}{(x - 2t)^2} \]

### Step 3: Differentiate the wave function with respect to time
To find the transverse velocity of a particle, we need to differentiate \( y \) with respect to \( t \):
\[ \frac{dy}{dt} = \frac{d}{dt} \left( \frac{6}{(x - 2t)^2} \right) \]
Using the chain rule, we get:
\[ \frac{dy}{dt} = 6 \cdot \frac{d}{dt} \left( (x - 2t)^{-2} \right) = 6 \cdot (-2)(x - 2t)^{-3} \cdot (-2) = \frac{24}{(x - 2t)^3} \]

### Step 4: Substitute the values of \( x \) and \( t \)
We need to find the transverse velocity at \( x = 2 \, \text{m} \) and \( t = 2 \, \text{s} \):
\[ \frac{dy}{dt} = \frac{24}{(2 - 2 \cdot 2)^3} = \frac{24}{(2 - 4)^3} = \frac{24}{(-2)^3} = \frac{24}{-8} = -3 \, \text{m/s} \]

### Step 5: Conclusion
The transverse velocity of the particle at \( x = 2 \, \text{m} \) and \( t = 2 \, \text{s} \) is:
\[ \frac{dy}{dt} = -3 \, \text{m/s} \]

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