A `4 kg` particle is moving along the x- axis under the action of the force `F = - ((pi^(2))/(16)) x N` At `t = 2 sec ` the particle passes through the origin and `t = 10sec`, the speed is `4sqrt(2)m//s` The amplitude of the motion is

Options

A. `(32sqrt2)/(pi)m`
B. `(16)/(pi)m`
C. `(4)/(pi)m`
D. `(16sqrt2)/(pi)m`

Explanation

To solve the problem step by step, we will follow the principles of simple harmonic motion (SHM) and the relationships between force, mass, angular frequency, displacement, and velocity.

### Step 1: Identify the Force and Its Relation to SHM
The force acting on the particle is given by:
\[ F = -\frac{\pi^2}{16} x \]
This can be compared to the standard form of Hooke's Law for SHM:
\[ F = -kx \]
From this, we can identify:
\[ k = \frac{\pi^2}{16} \]

### Step 2: Relate the Spring Constant to Angular Frequency
In SHM, the spring constant \( k \) is related to the mass \( m \) and angular frequency \( \omega \) by the formula:
\[ k = m \omega^2 \]
Given that the mass \( m = 4 \, \text{kg} \), we can substitute:
\[ \frac{\pi^2}{16} = 4 \omega^2 \]
Solving for \( \omega^2 \):
\[ \omega^2 = \frac{\pi^2}{16 \times 4} = \frac{\pi^2}{64} \]
Thus, taking the square root:
\[ \omega = \frac{\pi}{8} \]

### Step 3: Set Up the Equation of Motion
The general equation of motion for SHM is:
\[ x(t) = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude and \( \phi \) is the phase constant.

### Step 4: Use the Initial Condition
At \( t = 2 \, \text{s} \), the particle passes through the origin, so:
\[ x(2) = 0 \]
This gives us:
\[ A \sin(2\omega + \phi) = 0 \]
Since \( \sin \) is zero when its argument is a multiple of \( \pi \):
\[ 2\omega + \phi = n\pi \]
For simplicity, we can take \( n = 0 \):
\[ \phi = -2\omega = -2 \left(\frac{\pi}{8}\right) = -\frac{\pi}{4} \]

### Step 5: Use the Velocity Condition
The velocity \( v(t) \) in SHM is given by:
\[ v(t) = A \omega \cos(\omega t + \phi) \]
At \( t = 10 \, \text{s} \), the speed is given as \( 4\sqrt{2} \, \text{m/s} \):
\[ v(10) = 4\sqrt{2} = A \omega \cos(10\omega + \phi) \]
Substituting \( \omega = \frac{\pi}{8} \) and \( \phi = -\frac{\pi}{4} \):
\[ v(10) = A \left(\frac{\pi}{8}\right) \cos\left(10 \cdot \frac{\pi}{8} - \frac{\pi}{4}\right) \]
Calculating the argument of the cosine:
\[ 10 \cdot \frac{\pi}{8} - \frac{\pi}{4} = \frac{10\pi}{8} - \frac{2\pi}{8} = \frac{8\pi}{8} = \pi \]
Thus:
\[ \cos(\pi) = -1 \]
So:
\[ 4\sqrt{2} = A \left(\frac{\pi}{8}\right)(-1) \]
This simplifies to:
\[ A \frac{\pi}{8} = -4\sqrt{2} \]
Taking the absolute value:
\[ A \frac{\pi}{8} = 4\sqrt{2} \]
Solving for \( A \):
\[ A = \frac{4\sqrt{2} \cdot 8}{\pi} = \frac{32\sqrt{2}}{\pi} \]

### Final Answer
The amplitude of the motion is:
\[ A = \frac{32\sqrt{2}}{\pi} \, \text{meters} \]

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