A particle executes simple harmonic motion with frequqncy `2.5Hz` and amplitude `2 m`. The speed of the particle `0.3 s` after crossing, the equilibrium position is

Options

A. zero
B. `2pi m//s`
C. `4pi m//s`
D. `pi m//s`

Explanation

To find the speed of a particle executing simple harmonic motion (SHM) after a certain time has passed since it crossed the equilibrium position, we can follow these steps:

### Step 1: Understand the parameters given
- Frequency (f) = 2.5 Hz
- Amplitude (A) = 2 m
- Time after crossing equilibrium (t) = 0.3 s

### Step 2: Calculate the time period (T)
The time period (T) of SHM is the reciprocal of the frequency:
\[ T = \frac{1}{f} = \frac{1}{2.5 \text{ Hz}} = 0.4 \text{ s} \]

### Step 3: Determine the position of the particle at t = 0.3 s
In SHM, the particle moves from the equilibrium position to the extremes and back. The time taken for each quarter of the cycle is:
\[ \frac{T}{4} = \frac{0.4 \text{ s}}{4} = 0.1 \text{ s} \]
- From the equilibrium position to the right extreme takes 0.1 s.
- From the right extreme back to the equilibrium position takes another 0.1 s.
- From the equilibrium position to the left extreme takes another 0.1 s.

At \(t = 0.1 \text{ s}\), the particle reaches the right extreme. At \(t = 0.2 \text{ s}\), it returns to the equilibrium position. At \(t = 0.3 \text{ s}\), it reaches the left extreme.

### Step 4: Determine the speed at t = 0.3 s
At the extreme positions (left or right), the speed of the particle is zero. Therefore, the speed of the particle at \(t = 0.3 \text{ s}\) is:
\[ \text{Speed} = 0 \text{ m/s} \]

### Conclusion
The speed of the particle after crossing the equilibrium position at \(0.3 \text{ s}\) is \(0 \text{ m/s}\).

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