A wave moving with constant speed on a uniform string passes the point `x=0` with amplitude `A_0`, angular frequency `omega_0` and average rate of energy transfer `P_0`. As the wave travels down the string it gradually loses energy and at the point `x=l`, the average rate of energy transfer becomes`P_0`/2. At the point `x=l`. Angular frequency and amplitude are respectively:

Options

A. `omega_(0)"and "(A_(0))/(sqrt(2))`
B. `omega_(0)/(sqrt(2))"and"A_(0)`
C. less then `omega_(0)` and `A_(0)`
D. `(omega_(0))/(sqrt(2))` and `(A_(0)0/(sqrt(2))`

Answer

To solve the problem, we need to analyze how the amplitude and angular frequency of a wave change as it travels down a string while losing energy.

### Step-by-Step Solution:

1. **Understanding the Energy Transfer Relation**:
The average rate of energy transfer \( P \) for a wave on a string is given by the formula:
\[ P = \frac{1}{2} \mu A^2 \omega^2 V \]
where:
- \( \mu \) is the linear mass density of the string,
- \( A \) is the amplitude of the wave,
- \( \omega \) is the angular frequency,
- \( V \) is the speed of the wave.

2. **Initial Conditions**:
At the point \( x = 0 \):
- Amplitude = \( A_0 \)
- Angular frequency = \( \omega_0 \)
- Average rate of energy transfer = \( P_0 \)

Thus, we can write:
\[ P_0 = \frac{1}{2} \mu A_0^2 \omega_0^2 V \]

3. **Energy Transfer at Point \( x = l \)**:
At the point \( x = l \), the average rate of energy transfer becomes:
\[ P = \frac{P_0}{2} \]
Let the new amplitude at this point be \( A' \) (which we need to find), while \( \omega \) remains constant at \( \omega_0 \).

Therefore, we can write:
\[ \frac{P_0}{2} = \frac{1}{2} \mu (A')^2 \omega_0^2 V \]

4. **Setting Up the Equations**:
From the first equation:
\[ P_0 = \frac{1}{2} \mu A_0^2 \omega_0^2 V \]
From the second equation:
\[ \frac{P_0}{2} = \frac{1}{2} \mu (A')^2 \omega_0^2 V \]

5. **Dividing the Equations**:
Dividing the second equation by the first gives:
\[ \frac{\frac{P_0}{2}}{P_0} = \frac{\frac{1}{2} \mu (A')^2 \omega_0^2 V}{\frac{1}{2} \mu A_0^2 \omega_0^2 V} \]
Simplifying this, we get:
\[ \frac{1}{2} = \frac{(A')^2}{A_0^2} \]

6. **Solving for \( A' \)**:
Rearranging gives:
\[ (A')^2 = \frac{A_0^2}{2} \]
Taking the square root:
\[ A' = \frac{A_0}{\sqrt{2}} \]

7. **Angular Frequency**:
Since the angular frequency \( \omega \) remains constant, we have:
\[ \omega' = \omega_0 \]

### Final Result:
At the point \( x = l \):
- Amplitude \( A' = \frac{A_0}{\sqrt{2}} \)
- Angular frequency \( \omega' = \omega_0 \)