A perfectly elastic uniform string is suspended vertically with its upper end fixed to the ceiling and the lower end loaded with the weight. If a transverse wave is imparted to the lower end of the string, the pulse will

Options

A. not travel along the length of the string
B. travel upwards with increasing speed
C. travel upwards with decreasing speed
D. travelled upwards with constant acceleration

Explanation

To solve the problem, we need to analyze the behavior of a transverse wave on a vertically suspended string with a weight at the lower end. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Setup
We have a perfectly elastic uniform string that is fixed at one end to the ceiling and has a weight attached to the other end. This creates a tension in the string due to the weight.

### Step 2: Define the Variables
Let:
- \( m \) = mass of the weight attached at the lower end
- \( g \) = acceleration due to gravity
- \( \mu \) = linear density of the string (mass per unit length)
- \( T \) = tension in the string at a point \( x \) from the lower end
- \( x \) = distance from the lower end of the string

### Step 3: Calculate the Tension in the String
The tension \( T \) at a point \( x \) in the string can be expressed as:
\[ T = mg + \mu \cdot x \cdot g \]
This equation accounts for the weight of the mass and the weight of the portion of the string below the point \( x \).

### Step 4: Determine the Wave Velocity
The velocity \( V \) of the wave in the string is given by the formula:
\[ V = \sqrt{\frac{T}{\mu}} \]
Substituting the expression for tension \( T \):
\[ V = \sqrt{\frac{mg + \mu \cdot x \cdot g}{\mu}} = \sqrt{\frac{mg}{\mu} + x \cdot g} \]
This shows that the wave velocity \( V \) depends on the position \( x \) along the string.

### Step 5: Analyze the Dependency of Velocity on Position
From the equation derived, we see that as \( x \) increases (moving up the string), the wave velocity \( V \) also increases. This indicates a non-uniform wave speed along the string.

### Step 6: Determine the Acceleration
Using the relationship between acceleration \( a \), velocity \( V \), and position \( x \):
\[ a = V \frac{dV}{dx} \]
We can derive that the acceleration is constant:
\[ a = \frac{g}{2} \]
This indicates that the acceleration of the wave pulse is uniform along the string.

### Conclusion
Based on the analysis, we find that the pulse will exhibit characteristics of both increasing velocity with height and constant acceleration. Therefore, the correct options regarding the behavior of the wave pulse are options B and D.

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