The displacement from the position of equilibrium of a point 4 cm from a source of sinusoidal oscillations is half the amplitude at the moment `t=T//6` (T is the time perios). Assume that the source was at mean position at `t=0`. The wavelength of the running wave is :

A. 0.96 m
B. 0.48 m
C. 0.24 m
D. 0.12 m

Options

A. 0.96 m
B. 0.48 m
C. 0.24 m
D. 0.12 m

Explanation

To solve the problem, we need to find the wavelength of the running wave given the displacement from the equilibrium position at a certain distance and time. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the given information
- Displacement \( y \) at a distance \( x = 4 \) cm from the source is half the amplitude \( A \) at time \( t = \frac{T}{6} \).
- Thus, we can write:
\[ y = \frac{A}{2} \]

### Step 2: Use the wave equation
The general equation for a sinusoidal wave is given by:
\[ y = A \sin(\omega t - kx) \]
Where:
- \( \omega = \frac{2\pi}{T} \) is the angular frequency,
- \( k = \frac{2\pi}{\lambda} \) is the wave number.

### Step 3: Substitute the known values
At \( t = \frac{T}{6} \) and \( x = 4 \) cm (or \( 0.04 \) m), we substitute into the wave equation:
\[ \frac{A}{2} = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{6} - \frac{2\pi}{\lambda} \cdot 0.04\right) \]

### Step 4: Simplify the equation
Cancelling \( A \) from both sides (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin\left(\frac{\pi}{3} - \frac{8\pi}{\lambda}\right) \]

### Step 5: Solve for the sine value
We know that \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \). Thus, we can set up the equation:
\[ \frac{\pi}{3} - \frac{8\pi}{\lambda} = \frac{\pi}{6} \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ \frac{8\pi}{\lambda} = \frac{\pi}{3} - \frac{\pi}{6} \]
Calculating the right side:
\[ \frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi}{6} - \frac{\pi}{6} = \frac{\pi}{6} \]

### Step 7: Solve for \( \lambda \)
Now we have:
\[ \frac{8\pi}{\lambda} = \frac{\pi}{6} \]
Cross-multiplying gives:
\[ 8\pi \cdot 6 = \pi \lambda \]
Dividing both sides by \( \pi \):
\[ 48 = \lambda \]
Thus:
\[ \lambda = 0.48 \text{ m} \]

### Conclusion
The wavelength of the running wave is \( \lambda = 0.48 \) m, which corresponds to option B.

---