The equation for the displacement of a particle executing `SHM` is `y = 5 sin (2pit)cm`. Find
(i) velocity at `3cm` from the mean position
(ii) acceleration at `0.5s` after having the mean position

Options

A. `8pi cms^(-1), 0cms^(-2)`
B. `6pi cms^(-1), 20pi^(2) cms^(-2)`
C. `4pi cms^(-1), 0cms^(-2)`
D. `2pi cms^(-1), 20 pi^(2) cms^(-2)`

Explanation

To solve the problem, we will follow these steps:

### Given:
The displacement equation of the particle executing Simple Harmonic Motion (SHM) is:
\[ y = 5 \sin(2 \pi t) \text{ cm} \]

### (i) Finding the velocity at \( y = 3 \text{ cm} \):

1. **Differentiate the displacement equation to find the velocity**:
\[ v = \frac{dy}{dt} = \frac{d}{dt}[5 \sin(2 \pi t)] = 5 \cdot 2 \pi \cos(2 \pi t) = 10 \pi \cos(2 \pi t) \]

2. **Find \( t \) when \( y = 3 \text{ cm} \)**:
\[ 3 = 5 \sin(2 \pi t) \implies \sin(2 \pi t) = \frac{3}{5} \]

3. **Calculate \( \cos(2 \pi t) \)** using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ \cos(2 \pi t) = \sqrt{1 - \sin^2(2 \pi t)} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

4. **Substitute \( \cos(2 \pi t) \) back into the velocity equation**:
\[ v = 10 \pi \left(\frac{4}{5}\right) = 8 \pi \text{ cm/s} \]

### (ii) Finding the acceleration at \( t = 0.5 \text{ s} \):

1. **Differentiate the velocity equation to find the acceleration**:
\[ a = \frac{dv}{dt} = \frac{d}{dt}[10 \pi \cos(2 \pi t)] = 10 \pi \cdot (-2 \pi \sin(2 \pi t)) = -20 \pi^2 \sin(2 \pi t) \]

2. **Calculate \( \sin(2 \pi t) \) at \( t = 0.5 \text{ s} \)**:
\[ \sin(2 \pi \cdot 0.5) = \sin(\pi) = 0 \]

3. **Substitute \( \sin(2 \pi t) \) back into the acceleration equation**:
\[ a = -20 \pi^2 \cdot 0 = 0 \text{ cm/s}^2 \]

### Final Answers:
- (i) The velocity at \( 3 \text{ cm} \) from the mean position is \( 8 \pi \text{ cm/s} \).
- (ii) The acceleration at \( 0.5 \text{ s} \) after passing the mean position is \( 0 \text{ cm/s}^2 \).

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