A particle executing simple harmonic motion with an amplitude 5 cm and a time period 0.2 s. the velocity and acceleration of the particle when the displacement is 5 cm is

Options

A. `0.5pims^(-1),0ms^(-2)`
B. `0.5ms^(-1),-5pi^(2)ms^(-2)`
C. `0ms^(-1),-5pi^(2)ms^(-2)`
D. `0.5pims^(-1),-0.5pi^(2)ms^(-2)`

Explanation

To solve the problem of finding the velocity and acceleration of a particle executing simple harmonic motion (SHM) at a displacement of 5 cm, we can follow these steps:

### Step 1: Identify Given Values
- Amplitude (A) = 5 cm = 0.05 m (convert to meters for standard SI units)
- Time period (T) = 0.2 s
- Displacement (x) = 5 cm = 0.05 m

### Step 2: Calculate Angular Frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the time period:
\[ \omega = \frac{2\pi}{0.2} = 10\pi \, \text{rad/s} \]

### Step 3: Calculate Velocity (v)
The velocity of a particle in SHM is given by the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
Substituting the values:
\[ v = 10\pi \sqrt{(0.05)^2 - (0.05)^2} \]
\[ v = 10\pi \sqrt{0.0025 - 0.0025} = 10\pi \sqrt{0} = 0 \, \text{m/s} \]

### Step 4: Calculate Acceleration (a)
The acceleration of a particle in SHM is given by the formula:
\[ a = -\omega^2 x \]
Substituting the values:
\[ a = -(10\pi)^2 \cdot 0.05 \]
\[ a = -100\pi^2 \cdot 0.05 = -5\pi^2 \, \text{m/s}^2 \]

### Final Results
- Velocity (v) = 0 m/s
- Acceleration (a) = -5π² m/s²