If the maximum speed of a particle on a travelling wave is `v_(0)`, then find the speed of a particle when the displacement is half of the maximum value.

Options

A. `(v_(0))/(2)`
B. `sqrt(3v_(0))/(4)`
C. `sqrt(3 v_(0))/(2)`
D. `v_(0)`

Explanation

To solve the problem, we need to find the speed of a particle on a traveling wave when its displacement is half of the maximum value. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understand the Wave Equation**:
The standard equation for a traveling wave can be expressed as:
\[ y = A \sin(\omega t - kx) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, \( t \) is time, \( k \) is the wave number, and \( x \) is the position.

**Hint**: Familiarize yourself with the wave equation and its parameters.

2. **Find the Maximum Speed**:
The speed of a particle on the wave can be found by differentiating the wave equation with respect to time \( t \):
\[ v = \frac{dy}{dt} = A \omega \cos(\omega t - kx) \]
The maximum speed \( v_0 \) occurs when \( \cos(\omega t - kx) = 1 \):
\[ v_0 = A \omega \]

**Hint**: Remember that the maximum speed occurs when the cosine function equals 1.

3. **Determine Displacement Condition**:
We need to find the speed when the displacement \( y \) is half of the maximum value. Thus, we set:
\[ y = \frac{A}{2} \]
This implies:
\[ \frac{A}{2} = A \sin(\omega t - kx) \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin(\omega t - kx) \]

**Hint**: Use the sine function to relate displacement to the angle.

4. **Find the Angle**:
The equation \( \sin(\omega t - kx) = \frac{1}{2} \) corresponds to angles where:
\[ \omega t - kx = 30^\circ \quad \text{or} \quad \omega t - kx = 150^\circ \]
For simplicity, we can use \( 30^\circ \).

**Hint**: Recall the angles that yield a sine value of \( \frac{1}{2} \).

5. **Calculate the Speed at Half Displacement**:
Now, substitute \( \omega t - kx = 30^\circ \) back into the velocity equation:
\[ v = A \omega \cos(\omega t - kx) = A \omega \cos(30^\circ) \]
Since \( A \omega = v_0 \):
\[ v = v_0 \cos(30^\circ) \]
The value of \( \cos(30^\circ) \) is \( \frac{\sqrt{3}}{2} \):
\[ v = v_0 \cdot \frac{\sqrt{3}}{2} \]

**Hint**: Use the known value of cosine for \( 30^\circ \) to find the speed.

6. **Final Result**:
Thus, the speed of the particle when the displacement is half of the maximum value is:
\[ v = \frac{\sqrt{3}}{2} v_0 \]

### Conclusion:
The speed of a particle on a traveling wave when its displacement is half of the maximum value is \( \frac{\sqrt{3}}{2} v_0 \).