A transverse wave is passing through a stretched string with a speed of `20m//s`. The tension in the string is `20N`. At a certain point P on the string. It is observed that energy is being transferred at a rate of `40mW` at a given instant. Find the speed of point `P`.

Options

A. `40cm//s`
B. `20cm//s`
C. `2 mm//s`
D. `20mm//s`

Explanation

To solve the problem, we need to find the speed of point P on the string where the energy is being transferred at a rate of 40 mW. We will use the relationship between power, tension, wave speed, and particle speed in a transverse wave.

### Step 1: Understand the relationship between power, tension, and particle speed
The power \( P \) transmitted by a wave on a string can be expressed as:
\[ P = \frac{F \cdot V_p^2}{V} \]
where:
- \( P \) is the power (in watts),
- \( F \) is the tension in the string (in newtons),
- \( V_p \) is the particle speed at point P (in meters per second),
- \( V \) is the wave speed (in meters per second).

### Step 2: Substitute the known values into the equation
From the problem, we know:
- \( P = 40 \text{ mW} = 40 \times 10^{-3} \text{ W} \)
- \( F = 20 \text{ N} \)
- \( V = 20 \text{ m/s} \)

Substituting these values into the power equation gives:
\[ 40 \times 10^{-3} = \frac{20 \cdot V_p^2}{20} \]

### Step 3: Simplify the equation
The equation simplifies to:
\[ 40 \times 10^{-3} = V_p^2 \]

### Step 4: Solve for \( V_p \)
To find \( V_p \), we take the square root of both sides:
\[ V_p = \sqrt{40 \times 10^{-3}} = \sqrt{0.04} = 0.2 \text{ m/s} \]

### Step 5: Convert to centimeters per second
To convert meters per second to centimeters per second:
\[ V_p = 0.2 \text{ m/s} \times 100 = 20 \text{ cm/s} \]

### Final Answer
The speed of point P is \( 20 \text{ cm/s} \).

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