A particle executes simple harmonic motion about the point `x=0`. At time `t=0`, it has displacement `x=2cm` and zero velocity. If the frequency of motion is `0.25 s^(-1)`, find (a) the period, (b) angular frequency, ( c) the amplitude, (d) maximum speed, (e) the displacement from the mean position at `t=3 s`and (f) the velocity at `t=3 s`.

Explanation

To solve the problem step by step, we will use the principles of Simple Harmonic Motion (SHM).

### Given Data:
- Displacement at \( t = 0 \): \( x = 2 \, \text{cm} \)
- Velocity at \( t = 0 \): \( v = 0 \)
- Frequency \( f = 0.25 \, \text{s}^{-1} \)

### Step 1: Find the Period (T)
The period \( T \) is the reciprocal of the frequency \( f \):
\[ T = \frac{1}{f} = \frac{1}{0.25} = 4 \, \text{s} \]

### Step 2: Find the Angular Frequency (\( \omega \))
The angular frequency \( \omega \) is given by:
\[ \omega = 2\pi f = 2\pi \times 0.25 = \frac{\pi}{2} \, \text{rad/s} \]

### Step 3: Find the Amplitude (A)
The amplitude \( A \) is the maximum displacement from the mean position. Since the particle is at \( x = 2 \, \text{cm} \) at \( t = 0 \) and has zero velocity, this position is the maximum displacement:
\[ A = 2 \, \text{cm} \]

### Step 4: Find the Maximum Speed (\( v_{\text{max}} \))
The maximum speed in SHM is given by:
\[ v_{\text{max}} = A \cdot \omega = 2 \, \text{cm} \cdot \frac{\pi}{2} \, \text{rad/s} = \pi \, \text{cm/s} \]

### Step 5: Find the Displacement at \( t = 3 \, \text{s} \)
The displacement \( x(t) \) in SHM can be expressed as:
\[ x(t) = A \cos(\omega t + \phi) \]
Since the particle starts at the maximum displacement, \( \phi = 0 \) (cosine function is maximum at 0):
\[ x(3) = A \cos(\omega \cdot 3) = 2 \cos\left(\frac{\pi}{2} \cdot 3\right) = 2 \cos\left(\frac{3\pi}{2}\right) \]
Since \( \cos\left(\frac{3\pi}{2}\right) = 0 \):
\[ x(3) = 2 \cdot 0 = 0 \, \text{cm} \]

### Step 6: Find the Velocity at \( t = 3 \, \text{s} \)
The velocity \( v(t) \) in SHM can be expressed as:
\[ v(t) = -A \omega \sin(\omega t + \phi) \]
Substituting the values:
\[ v(3) = -2 \cdot \frac{\pi}{2} \sin\left(\frac{\pi}{2} \cdot 3\right) = -\pi \sin\left(\frac{3\pi}{2}\right) \]
Since \( \sin\left(\frac{3\pi}{2}\right) = -1 \):
\[ v(3) = -\pi \cdot (-1) = \pi \, \text{cm/s} \]

### Summary of Results:
(a) Period \( T = 4 \, \text{s} \)
(b) Angular Frequency \( \omega = \frac{\pi}{2} \, \text{rad/s} \)
(c) Amplitude \( A = 2 \, \text{cm} \)
(d) Maximum Speed \( v_{\text{max}} = \pi \, \text{cm/s} \)
(e) Displacement at \( t = 3 \, \text{s} \): \( x(3) = 0 \, \text{cm} \)
(f) Velocity at \( t = 3 \, \text{s} \): \( v(3) = \pi \, \text{cm/s} \)