The equation `vec(phi)(x,t)=vec(j)sin ((2pi)/(lambda)vt)cos ((2 pi)/(lambda)x)` represents

Options

A. transverse progressive wave
B. longitudinal progressive
C. transverse stationary wave
D. none of the above

Answer

To analyze the given equation \(\vec{\phi}(x,t) = \vec{j} \sin\left(\frac{2\pi}{\lambda} vt\right) \cos\left(\frac{2\pi}{\lambda} x\right)\), we will determine the type of wave it represents step by step.

### Step 1: Identify the Components of the Wave Equation
The equation consists of two parts:
- A sine function \(\sin\left(\frac{2\pi}{\lambda} vt\right)\) which depends on time \(t\).
- A cosine function \(\cos\left(\frac{2\pi}{\lambda} x\right)\) which depends on position \(x\).

### Step 2: Determine the Nature of the Wave
To classify the wave as stationary or progressive, we need to check if there are points where the amplitude is always zero (nodal points).

### Step 3: Find Nodal Points
The amplitude of the wave is given by the product of the sine and cosine functions. The amplitude will be zero when:
\[ \sin\left(\frac{2\pi}{\lambda} vt\right) = 0 \]
This occurs at:
\[ \frac{2\pi}{\lambda} vt = n\pi \quad \text{for } n = 0, 1, 2, \ldots \]
Thus, we can express this as:
\[ t = \frac{n\lambda}{2v} \]

### Step 4: Analyze the Cosine Component
The cosine function \(\cos\left(\frac{2\pi}{\lambda} x\right)\) will also be zero at certain points:
\[ \cos\left(\frac{2\pi}{\lambda} x\right) = 0 \]
This occurs at:
\[ \frac{2\pi}{\lambda} x = \frac{\pi}{2} + m\pi \quad \text{for } m = 0, 1, 2, \ldots \]
Thus, we can express this as:
\[ x = \frac{(2m+1)\lambda}{4} \]

### Step 5: Conclusion on the Type of Wave
Since both the sine and cosine components yield points where the amplitude is zero, we conclude that the wave has nodal points. Therefore, it is classified as a stationary wave.

### Step 6: Identify the Direction of Propagation and Vibration
The wave propagates in the \(x\) direction (as indicated by the \(x\) term) and the vibration occurs in the \(y\) direction (as indicated by the \(\vec{j}\) term). This confirms that the wave is a transverse wave.

### Final Conclusion
The given equation represents a **transverse stationary wave**.

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