A particle executing SHM of amplitude `a` has displace ment `(a)/(2)` at `t=(T)/(4)` and a negative velocity. The epoch of the particle is

Options

A. `(pi)/(3)`
B. `(2pi)/(3)`
C. `pi`
D. `(5pi)/(3)`

Explanation

To solve the problem step by step, we will analyze the given information about the particle executing Simple Harmonic Motion (SHM).

### Step 1: Understand the displacement equation of SHM
The displacement \( x \) of a particle in SHM can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the initial phase (epoch).

### Step 2: Substitute the known values
We know that at \( t = \frac{T}{4} \), the displacement \( x \) is \( \frac{a}{2} \) (where \( A = a \)):
\[ \frac{a}{2} = a \sin\left(\omega \frac{T}{4} + \phi\right) \]
Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ \frac{1}{2} = \sin\left(\omega \frac{T}{4} + \phi\right) \]

### Step 3: Determine the angular frequency
The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into the equation:
\[ \frac{1}{2} = \sin\left(\frac{2\pi}{T} \cdot \frac{T}{4} + \phi\right) \]
This simplifies to:
\[ \frac{1}{2} = \sin\left(\frac{\pi}{2} + \phi\right) \]

### Step 4: Solve for the phase \( \phi \)
The sine function equals \( \frac{1}{2} \) at two angles:
\[ \frac{\pi}{6} \quad \text{and} \quad \frac{5\pi}{6} \]
Thus, we have:
\[ \frac{\pi}{2} + \phi = \frac{\pi}{6} \quad \text{or} \quad \frac{\pi}{2} + \phi = \frac{5\pi}{6} \]
From these equations, we can solve for \( \phi \):
1. For \( \frac{\pi}{2} + \phi = \frac{\pi}{6} \):
\[ \phi = \frac{\pi}{6} - \frac{\pi}{2} = -\frac{\pi}{3} \]
2. For \( \frac{\pi}{2} + \phi = \frac{5\pi}{6} \):
\[ \phi = \frac{5\pi}{6} - \frac{\pi}{2} = \frac{5\pi}{6} - \frac{3\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

### Step 5: Analyze the velocity condition
The velocity \( v \) in SHM is given by:
\[ v(t) = A \omega \cos(\omega t + \phi) \]
At \( t = \frac{T}{4} \):
\[ v\left(\frac{T}{4}\right) = A \omega \cos\left(\frac{\pi}{2} + \phi\right) \]
This simplifies to:
\[ v\left(\frac{T}{4}\right) = A \omega \cos\left(\phi + \frac{\pi}{2}\right) = -A \omega \sin(\phi) \]
We know from the problem that the velocity is negative, which means:
\[ -A \omega \sin(\phi) < 0 \implies \sin(\phi) > 0 \]

### Step 6: Determine the correct phase
1. For \( \phi = \frac{\pi}{3} \):
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} > 0 \quad \text{(valid)} \]
2. For \( \phi = -\frac{\pi}{3} \):
\[ \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} < 0 \quad \text{(not valid)} \]

### Conclusion
Since only \( \phi = \frac{\pi}{3} \) satisfies the condition of having a negative velocity, the epoch of the particle is:
\[ \phi = \frac{\pi}{3} \]