A pulse is started at a time `t = 0` along the `+x` directions an a long, taut string. The shaot of the puise at `t = 0` is given by funcation `y` with
`y = {{:((x)/(4)+1fo r-4ltxle0),(-x+1fo r0ltxlt1),("0 otherwise"):}`
here `y` and `x` are in centimeters. The linear mass density of the string is `50 g//m` and it is under a tension of `5N`,
The transverse velocity of the particle at `x = 13 cm` and `t = 0.015 s` will be

Options

A. `-250 cm//s`
B. `-500 cm//s`
C. `500 cm//s`
D. `-1000 cm//s`

Explanation

To solve the problem, we need to find the transverse velocity of the particle at \( x = 13 \, \text{cm} \) and \( t = 0.015 \, \text{s} \). Let's break down the solution step by step:

### Step 1: Understand the Wave Equation
The pulse is defined by the function \( y \) at \( t = 0 \):
- \( y = \frac{x}{4} + 1 \) for \( -4 < x \leq 0 \)
- \( y = -x + 1 \) for \( 0 < x < 1 \)
- \( y = 0 \) otherwise

### Step 2: Determine the Wave Velocity
The wave velocity \( v \) can be calculated using the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension and \( \mu \) is the linear mass density.

Given:
- \( T = 5 \, \text{N} \)
- \( \mu = 50 \, \text{g/m} = 0.05 \, \text{kg/m} \)

Substituting these values:
\[ v = \sqrt{\frac{5}{0.05}} = \sqrt{100} = 10 \, \text{m/s} = 1000 \, \text{cm/s} \]

### Step 3: Find the Slope of the Pulse
To find the transverse velocity, we need to calculate the slope \( \frac{dy}{dx} \) of the wave function at the given position.

At \( x = 13 \, \text{cm} \) and \( t = 0.015 \, \text{s} \), the wave has already moved past the initial pulse defined by \( y \). Since \( x = 13 \, \text{cm} \) is outside the defined pulse region, we need to determine the slope of the wave at the edge of the pulse.

The slope of the line \( y = \frac{x}{4} + 1 \) is \( \frac{dy}{dx} = \frac{1}{4} \) for \( -4 < x \leq 0 \) and the slope of the line \( y = -x + 1 \) is \( \frac{dy}{dx} = -1 \) for \( 0 < x < 1 \).

Since \( x = 13 \, \text{cm} \) is outside the range of the pulse, we assume the slope is effectively \( 0 \) because the wave has already passed that point.

### Step 4: Calculate the Transverse Velocity
The transverse velocity \( v_p \) of the particle is given by:
\[ v_p = -\frac{dy}{dx} \cdot v \]

Since \( \frac{dy}{dx} \) is \( 0 \) at \( x = 13 \, \text{cm} \):
\[ v_p = -0 \cdot 1000 \, \text{cm/s} = 0 \, \text{cm/s} \]

### Conclusion
Thus, the transverse velocity of the particle at \( x = 13 \, \text{cm} \) and \( t = 0.015 \, \text{s} \) is \( 0 \, \text{cm/s} \).