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. `(32sqrt(2))/(pi)m`
B. `(16)/(pi)m`
C. `(4)/(pi)m`
D. `(16sqrt(2))/(pi)m`

Explanation

To solve the problem step by step, we will analyze the given force, relate it to simple harmonic motion, and find the amplitude of the motion.

### Step 1: Identify the force and relate it to simple harmonic motion
The force acting on the particle is given by:
\[ F = -\frac{\pi^2}{16} x \]

This is in the form of Hooke's law, \( F = -kx \), where \( k = \frac{\pi^2}{16} \). In simple harmonic motion, we have:
\[ F = m \frac{d^2x}{dt^2} \]
Thus, we can equate:
\[ m \omega^2 = k \]
where \( \omega \) is the angular frequency.

### Step 2: Calculate the angular frequency \( \omega \)
Given the mass \( m = 4 \, \text{kg} \):
\[ \omega^2 = \frac{k}{m} = \frac{\frac{\pi^2}{16}}{4} = \frac{\pi^2}{64} \]
Taking the square root to find \( \omega \):
\[ \omega = \frac{\pi}{8} \, \text{rad/s} \]

### Step 3: Write the equation of motion
The general solution for the position \( x \) in simple harmonic motion 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 \( t = 2 \, \text{s} \)
At \( t = 2 \, \text{s} \), the particle passes through the origin, so:
\[ x(2) = A \sin\left(\frac{\pi}{8} \cdot 2 + \phi\right) = 0 \]
This implies:
\[ \sin\left(\frac{\pi}{4} + \phi\right) = 0 \]
Thus:
\[ \frac{\pi}{4} + \phi = n\pi \quad \text{for some integer } n \]
Choosing \( n = 0 \):
\[ \phi = -\frac{\pi}{4} \]

### Step 5: Write the position equation with the phase constant
Now substituting \( \phi \) back into the equation:
\[ x(t) = A \sin\left(\frac{\pi}{8} t - \frac{\pi}{4}\right) \]

### Step 6: Use the condition at \( t = 10 \, \text{s} \) to find the amplitude
The speed \( v(t) \) is given by the derivative of \( x(t) \):
\[ v(t) = \frac{dx}{dt} = A \omega \cos\left(\frac{\pi}{8} t - \frac{\pi}{4}\right) \]
At \( t = 10 \, \text{s} \), the speed is given as \( 4\sqrt{2} \):
\[ 4\sqrt{2} = A \cdot \frac{\pi}{8} \cos\left(\frac{\pi}{8} \cdot 10 - \frac{\pi}{4}\right) \]
Calculating the argument:
\[ \frac{\pi}{8} \cdot 10 - \frac{\pi}{4} = \frac{5\pi}{4} - \frac{2\pi}{4} = \frac{3\pi}{4} \]
Thus:
\[ \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}} \]
Substituting back:
\[ 4\sqrt{2} = A \cdot \frac{\pi}{8} \left(-\frac{1}{\sqrt{2}}\right) \]
Rearranging gives:
\[ A = \frac{4\sqrt{2} \cdot 8}{-\pi/\sqrt{2}} = -\frac{32}{\pi} \]
Taking the absolute value (since amplitude is positive):
\[ A = \frac{32}{\pi} \]

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