The phase of a particle executing simple harmonic motion is `pi/2` when it has

Options

A. Maximum velocity
B. Maximum acceleration
C. Maximum energy
D. Maximum displacement

Answer

To solve the problem regarding the phase of a particle executing simple harmonic motion (SHM) when it is at `π/2`, we can follow these steps:

### Step 1: Understand the Equation of SHM
The general equation for the displacement \( x \) of a particle in SHM is given by:
\[ x(t) = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( t \) is the time,
- \( \phi \) is the phase constant.

### Step 2: Set the Phase to \( \pi/2 \)
Given that the phase is \( \pi/2 \), we can set:
\[ \omega t + \phi = \frac{\pi}{2} \]
Assuming the phase constant \( \phi = 0 \) for simplicity, we have:
\[ \omega t = \frac{\pi}{2} \]

### Step 3: Solve for Time \( t \)
From the equation \( \omega t = \frac{\pi}{2} \), we can solve for \( t \):
\[ t = \frac{\pi}{2\omega} \]

### Step 4: Relate Angular Frequency to Time Period
The angular frequency \( \omega \) is related to the time period \( T \) by the equation:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into our equation for \( t \):
\[ t = \frac{\pi}{2 \cdot \frac{2\pi}{T}} = \frac{T}{4} \]

### Step 5: Interpret the Result
The result \( t = \frac{T}{4} \) indicates that the particle has traveled for one-fourth of its time period. In the context of SHM, this means that at \( t = \frac{T}{4} \), the particle is at its maximum displacement (amplitude \( A \)).

### Step 6: Conclusion
Thus, when the phase of a particle executing SHM is \( \pi/2 \), it is at its maximum displacement (amplitude).

### Final Answer
The phase of a particle executing simple harmonic motion is \( \pi/2 \) when it has reached its maximum displacement (amplitude).

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