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 than `omega_(0)` and `A_(0)`
D. `omega_(0)//sqrt(2)` and `A_(0)//sqrt(2)`

Explanation

To solve the problem step by step, we will analyze the energy transfer in the wave on the string and how it relates to amplitude and angular frequency.

### Step 1: Understand the initial conditions
The wave at point \( x = 0 \) has:
- Amplitude: \( A_0 \)
- Angular frequency: \( \omega_0 \)
- Average rate of energy transfer: \( P_0 \)

### Step 2: Write the expression for energy transfer
The average power (rate of energy transfer) for a wave on a string can be expressed as:
\[ P = \frac{1}{2} \mu \omega^2 A^2 v \]
where:
- \( \mu \) = mass per unit length of the string
- \( \omega \) = angular frequency
- \( A \) = amplitude
- \( v \) = wave speed

### Step 3: Write the initial power equation
At \( x = 0 \):
\[ P_0 = \frac{1}{2} \mu \omega_0^2 A_0^2 v \]

### Step 4: Write the power equation at \( x = l \)
At \( x = l \), the power is given as \( \frac{P_0}{2} \):
\[ \frac{P_0}{2} = \frac{1}{2} \mu \omega^2 A^2 v \]

### Step 5: Relate the two power equations
From the two equations, we can relate the powers:
\[ \frac{P_0}{\frac{P_0}{2}} = \frac{\frac{1}{2} \mu \omega_0^2 A_0^2 v}{\frac{1}{2} \mu \omega^2 A^2 v} \]
This simplifies to:
\[ 2 = \frac{\omega_0^2 A_0^2}{\omega^2 A^2} \]

### Step 6: Isolate the variables
Rearranging gives:
\[ \omega^2 A^2 = \frac{A_0^2}{2} \omega_0^2 \]

### Step 7: Find the amplitude at \( x = l \)
We can express the amplitude \( A \) at \( x = l \):
\[ A^2 = \frac{A_0^2}{2} \]
Taking the square root:
\[ A = \frac{A_0}{\sqrt{2}} \]

### Step 8: Determine the angular frequency
Since the angular frequency depends on the source and is not affected by the energy loss in the wave, we have:
\[ \omega = \omega_0 \]

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

### Summary of Results
- Amplitude at \( x = l \): \( A = \frac{A_0}{\sqrt{2}} \)
- Angular frequency at \( x = l \): \( \omega = \omega_0 \)