A particle executing SHM along y-axis, which is described by `y = 10 "sin"(pi t)/(4)`, phase of particle at t =2s is

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. `(pi)/(8)`
D. `pi`

Answer

To find the phase of a particle executing simple harmonic motion (SHM) described by the equation \( y = 10 \sin\left(\frac{\pi t}{4}\right) \) at \( t = 2 \) seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the General Form of SHM**:
The general equation for SHM is given by:
\[ y = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( t \) is time,
- \( \phi \) is the initial phase.

2. **Extract the Phase Expression**:
From the given equation \( y = 10 \sin\left(\frac{\pi t}{4}\right) \), we can see that:
- The angular frequency \( \omega = \frac{\pi}{4} \).
- The initial phase \( \phi = 0 \) (since there is no additional phase term).

3. **Determine the Phase at Time \( t = 2 \) seconds**:
The phase of the particle at any time \( t \) can be expressed as:
\[ \text{Phase} = \omega t + \phi \]
Substituting the values we have:
\[ \text{Phase} = \left(\frac{\pi}{4}\right) \cdot 2 + 0 \]

4. **Calculate the Phase**:
Now, we calculate:
\[ \text{Phase} = \frac{\pi \cdot 2}{4} = \frac{2\pi}{4} = \frac{\pi}{2} \]

5. **Final Answer**:
Therefore, the phase of the particle at \( t = 2 \) seconds is:
\[ \frac{\pi}{2} \]

### Summary:
The phase of the particle at \( t = 2 \) seconds is \( \frac{\pi}{2} \).

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