A standing wave of time period `T` is set up in a string clamped between two rigid supports, At `t =0` antinode is at its maximum displacement `2A`.

Options

A. The energy density of a node is equal to energy density of an antinode for the first time at `t= T//4`
B. The energy density of node and antinode becomes equal after `T//2` second
C. The displacement of the particle at antinode at `t = (T)/(8)` is `sqrt(2) A`
D. The displacement of the particle at node is zero

Explanation

To solve the problem of a standing wave in a string clamped between two rigid supports, we will analyze the situation step by step.

### Step 1: Understand the standing wave equation
The displacement of a standing wave can be described by the equation:
\[ y(x, t) = 2A \sin(kx) \cos(\omega t) \]
where:
- \( A \) is the amplitude,
- \( k \) is the wave number,
- \( \omega \) is the angular frequency,
- \( t \) is time.

### Step 2: Identify the parameters
Given that at \( t = 0 \), the antinode is at its maximum displacement of \( 2A \), we can deduce that:
- At \( t = 0 \), \( \cos(\omega t) = 1 \) (since \( \cos(0) = 1 \)).
- Therefore, the displacement at the antinode is:
\[ y(x, 0) = 2A \sin(kx) \]

### Step 3: Determine the position of the antinode
The antinodes occur where \( \sin(kx) = 1 \). This means:
\[ kx = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
For the first antinode, we can take \( n = 0 \):
\[ kx = \frac{\pi}{2} \]
Thus, the position of the first antinode is at:
\[ x = \frac{\pi}{2k} \]

### Step 4: Analyze the displacement at \( t = \frac{T}{8} \)
Now, we need to find the displacement at \( t = \frac{T}{8} \). The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Substituting \( t = \frac{T}{8} \) into the wave equation:
\[ y(x, \frac{T}{8}) = 2A \sin(kx) \cos\left(\frac{2\pi}{T} \cdot \frac{T}{8}\right) \]
This simplifies to:
\[ y(x, \frac{T}{8}) = 2A \sin(kx) \cos\left(\frac{\pi}{4}\right) \]
Since \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):
\[ y(x, \frac{T}{8}) = 2A \sin(kx) \cdot \frac{1}{\sqrt{2}} = \frac{2A}{\sqrt{2}} \sin(kx) = \sqrt{2}A \sin(kx) \]

### Step 5: Evaluate the displacement at the antinode
At the antinode (where \( \sin(kx) = 1 \)):
\[ y(x, \frac{T}{8}) = \sqrt{2}A \]

### Conclusion
Thus, the displacement of the particle at the antinode at \( t = \frac{T}{8} \) is:
\[ y(x, \frac{T}{8}) = \sqrt{2}A \]

### Final Answer
The displacement of the particle at the antinode at \( t = \frac{T}{8} \) is \( \sqrt{2}A \).

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