A pulse is started at a time `t=0` along the `+x`direction on a long, taut string. The shape of the pulse at `t=0` is given by function `f(x)` with
here `f` 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. (a)`-250 cm//s`
B. (b)`-500 cm//s`
C. (c)`500 cm//s`
D. (d)`-1000 cm//s`

Explanation

To solve the problem, we need to find the transverse velocity of a particle on the string at a given position and time. Let's break it down step by step.

### Step 1: Identify the parameters given in the problem.
- Linear mass density of the string, \( \mu = 50 \, \text{g/m} = 0.05 \, \text{kg/m} \) (since \( 1 \, \text{g} = 0.001 \, \text{kg} \))
- Tension in the string, \( T = 5 \, \text{N} \)
- Position, \( x = 13 \, \text{cm} = 0.13 \, \text{m} \)
- Time, \( t = 0.015 \, \text{s} \)

### Step 2: Calculate the wave speed \( v \) on the string.
The wave speed \( v \) on a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the values:
\[ v = \sqrt{\frac{5 \, \text{N}}{0.05 \, \text{kg/m}}} = \sqrt{100} = 10 \, \text{m/s} \]

### Step 3: Determine the distance traveled by the wave in the given time.
The distance traveled by the wave in time \( t \) is:
\[ \text{Distance} = v \cdot t = 10 \, \text{m/s} \cdot 0.015 \, \text{s} = 0.15 \, \text{m} = 15 \, \text{cm} \]

### Step 4: Analyze the position of the particle.
The wave pulse starts at \( x = 0 \) and travels in the positive x-direction. After \( t = 0.015 \, \text{s} \), the wave has traveled \( 15 \, \text{cm} \).

### Step 5: Determine if the particle at \( x = 13 \, \text{cm} \) is within the wave pulse.
Since the wave has traveled \( 15 \, \text{cm} \) and the particle is at \( x = 13 \, \text{cm} \), it is indeed within the wave pulse.

### Step 6: Calculate the transverse velocity \( \frac{dy}{dt} \).
The transverse velocity of the particle can be calculated using the relationship:
\[ \frac{dy}{dt} = -\frac{v}{4} \]
Here, we use \( v = 10 \, \text{m/s} \):
\[ \frac{dy}{dt} = -\frac{10 \, \text{m/s}}{4} = -2.5 \, \text{m/s} = -250 \, \text{cm/s} \]

### Conclusion
The transverse velocity of the particle at \( x = 13 \, \text{cm} \) and \( t = 0.015 \, \text{s} \) is \( -250 \, \text{cm/s} \).