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, we need to find the velocity and acceleration of a particle executing simple harmonic motion (SHM) when its displacement is equal to the amplitude. Here are the steps to derive the solution:

### Step 1: Identify the Given Values
- Amplitude (A) = 5 cm = 0.05 m (convert to meters)
- Time Period (T) = 0.2 s

### Step 2: Calculate Angular Frequency (ω)
The angular frequency (ω) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the given time period:
\[ \omega = \frac{2\pi}{0.2} = 10\pi \, \text{radians/second} \]

### Step 3: Determine the Velocity (v) at Displacement (x = A)
In SHM, the velocity can be calculated using the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
Since we are finding the velocity when the displacement (x) is equal to the amplitude (A), we have:
\[ x = A = 0.05 \, \text{m} \]
Substituting the values into the velocity formula:
\[ v = 10\pi \sqrt{(0.05)^2 - (0.05)^2} \]
This simplifies to:
\[ v = 10\pi \sqrt{0} = 0 \, \text{m/s} \]

### Step 4: Calculate Acceleration (a) at Displacement (x = A)
The acceleration in SHM can be calculated using the formula:
\[ a = -\omega^2 x \]
Substituting the values:
\[ a = - (10\pi)^2 (0.05) \]
Calculating this gives:
\[ a = -100\pi^2 \times 0.05 = -5\pi^2 \, \text{m/s}^2 \]

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

### Summary
- When the displacement is equal to the amplitude (5 cm), the velocity of the particle is 0 m/s, and the acceleration is -5π² m/s².