An oscillator of mass M is at rest in its equilibrium position in a potential `V=(1)/(2)k(x-X)^(2)`. A particle of massm comes from right with speed u and collides completely inelastically with M and sticks to it. The process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscilllations after 13 collisions is : (M=10, m=5, u=1 ,k=1)

Options

A. `sqrt((3)/(5))`
B. `(2)/(3)`
C. `(1)/(sqrt3)`
D. `(1)/(2)`

Answer

To solve the problem step-by-step, we will analyze the situation involving an oscillator and a particle colliding with it.

### Step 1: Understand the System
We have an oscillator of mass \( M \) (10 kg) at rest in its equilibrium position described by the potential energy function \( V = \frac{1}{2} k (x - X)^2 \). A particle of mass \( m \) (5 kg) comes in with speed \( u \) (1 m/s) and collides completely inelastically with the oscillator.

### Step 2: Conservation of Momentum
Since the collision is completely inelastic, we can use the conservation of momentum. The momentum before the collision is equal to the momentum after the collision.

Before the collision:
\[ p_{\text{initial}} = m \cdot u = 5 \cdot 1 = 5 \, \text{kg m/s} \]

After the collision, the combined mass is \( M + m \):
\[ p_{\text{final}} = (M + m) \cdot v \]
where \( v \) is the velocity after the collision.

Setting the initial momentum equal to the final momentum:
\[ 5 = (10 + 5) \cdot v \implies 5 = 15v \implies v = \frac{1}{3} \, \text{m/s} \]

### Step 3: Kinetic Energy After Collision
The kinetic energy of the system after the collision can be calculated using:
\[ KE = \frac{1}{2} (M + m) v^2 \]
Substituting the values:
\[ KE = \frac{1}{2} (10 + 5) \left(\frac{1}{3}\right)^2 = \frac{1}{2} \cdot 15 \cdot \frac{1}{9} = \frac{15}{18} = \frac{5}{6} \, \text{J} \]

### Step 4: Amplitude of Oscillation
The kinetic energy at the mean position is related to the amplitude \( A \) of the oscillation by:
\[ KE = \frac{1}{2} k A^2 \]
Given \( k = 1 \):
\[ \frac{5}{6} = \frac{1}{2} A^2 \implies A^2 = \frac{5}{3} \implies A = \sqrt{\frac{5}{3}} \, \text{m} \]

### Step 5: Mass After Each Collision
After each collision, the mass of the oscillator increases by \( m \). After \( n \) collisions, the mass \( M_n \) becomes:
\[ M_n = M + n \cdot m = 10 + n \cdot 5 \]
For \( n = 13 \):
\[ M_{13} = 10 + 13 \cdot 5 = 10 + 65 = 75 \, \text{kg} \]

### Step 6: Kinetic Energy After 13 Collisions
The new kinetic energy after 13 collisions can be calculated using the new mass:
\[ p = 5 \, \text{kg m/s} \quad \text{(momentum remains constant)} \]
\[ KE = \frac{p^2}{2M_n} = \frac{25}{2 \cdot 75} = \frac{25}{150} = \frac{1}{6} \, \text{J} \]

### Step 7: Amplitude After 13 Collisions
Using the kinetic energy to find the new amplitude:
\[ \frac{1}{6} = \frac{1}{2} A_{13}^2 \implies A_{13}^2 = \frac{1}{3} \implies A_{13} = \frac{1}{\sqrt{3}} \, \text{m} \]

### Final Answer
The amplitude of oscillations after 13 collisions is:
\[ A_{13} = \frac{1}{\sqrt{3}} \, \text{m} \]