A particle performing `SHM` is found at its eqilbrium at `t=1 sec`.and it is found to have a speed of `0.25 m//s` at `t=2sec`. If the period of oscillation is `6sec`. Calculate amplitude of oscillation-

Options

A. `(2)/(2pi)m`
B. `(3)/(2pi)m`
C. `(6)/(pi)m`
D. `(3)/(8pi)`

Answer

To solve the problem step by step, we will follow the process of finding the amplitude of the particle performing Simple Harmonic Motion (SHM) using the given information.

### Step 1: Understand the Given Information
- The particle is at equilibrium (mean position) at \( t = 1 \) sec.
- The speed of the particle at \( t = 2 \) sec is \( 0.25 \, \text{m/s} \).
- The period of oscillation \( T = 6 \) sec.

### Step 2: Calculate Angular Frequency (\( \omega \))
The angular frequency \( \omega \) is related to the period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{6} = \frac{\pi}{3} \, \text{rad/s} \]

### Step 3: Write the Equations for Displacement and Velocity
The displacement \( x(t) \) and velocity \( v(t) \) for SHM can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
\[ v(t) = A \omega \cos(\omega t + \phi) \]

### Step 4: Use the Information at \( t = 1 \) sec
At \( t = 1 \) sec, the particle is at equilibrium, which means:
\[ x(1) = 0 \]
Substituting into the displacement equation:
\[ 0 = A \sin\left(\frac{\pi}{3} \cdot 1 + \phi\right) \]
This implies:
\[ \sin\left(\frac{\pi}{3} + \phi\right) = 0 \]
The sine function is zero when its argument is \( n\pi \) (where \( n \) is an integer). For the first oscillation, we can take \( n = 1 \):
\[ \frac{\pi}{3} + \phi = \pi \implies \phi = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

### Step 5: Use the Information at \( t = 2 \) sec
Now, we can use the speed at \( t = 2 \) sec:
\[ v(2) = 0.25 \, \text{m/s} \]
Substituting into the velocity equation:
\[ 0.25 = A \cdot \frac{\pi}{3} \cos\left(\frac{\pi}{3} \cdot 2 + \frac{2\pi}{3}\right) \]
Calculating the argument of the cosine:
\[ \frac{\pi}{3} \cdot 2 + \frac{2\pi}{3} = \frac{2\pi}{3} + \frac{2\pi}{3} = \frac{4\pi}{3} \]
Now substituting into the cosine function:
\[ \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \]
Substituting this back into the velocity equation:
\[ 0.25 = A \cdot \frac{\pi}{3} \cdot \left(-\frac{1}{2}\right) \]
This simplifies to:
\[ 0.25 = -\frac{A\pi}{6} \]
Solving for \( A \):
\[ A = -\frac{0.25 \cdot 6}{\pi} = \frac{1.5}{\pi} \]

### Step 6: Final Result
The amplitude \( A \) of the oscillation is:
\[ A = \frac{3}{2\pi} \, \text{m} \]