One end of an ideal spring is fixed to a wall at origin `O` and axis of spring is parallel to x-axis. A block of mass `m = 1kg` is attached to free end of the spring and it is performing SHM. Equation of position of the block in co-ordinate system shown in figure is `x = 10 + 3 sin (10t)`. Here, t is in second and `x` in `cm`. Another block of mass `M = 3kg`, moving towards the origin with velocity `30 cm//s` collides with the block performing SHM at `t = 0` and gets stuck to it. Calculate
(a) new amplitude of oscillations,
(b) new equation for position of the combined body,
( c) loss of energy during collision. Neglect friction.

Options

A. 20 `rad//s`
B. 5 `rad//s`
C. 100`rad//s`
D. 50`rad//s`

Explanation

To solve the problem step by step, we will break down the calculations for each part of the question.

### Given Data:
- Mass of block \( m = 1 \, \text{kg} \)
- Mass of block \( M = 3 \, \text{kg} \)
- Initial position equation of the block performing SHM:
\[ x = 10 + 3 \sin(10t) \quad (\text{in cm}) \]
- Velocity of block \( M \) before collision: \( v_M = 30 \, \text{cm/s} = 0.3 \, \text{m/s} \)

### Part (a): New Amplitude of Oscillations

1. **Calculate the initial momentum before the collision:**
- The momentum of block \( m \) (1 kg) at \( t = 0 \):
\[ v_m = \frac{dx}{dt} = 3 \cdot 10 \cos(10 \cdot 0) = 30 \, \text{cm/s} = 0.3 \, \text{m/s} \]
- The momentum of block \( M \) (3 kg):
\[ p_M = M \cdot v_M = 3 \cdot 0.3 = 0.9 \, \text{kg m/s} \]
- Total initial momentum:
\[ p_{\text{initial}} = p_m + p_M = (1 \cdot 0.3) + (3 \cdot (-0.3)) = 0.3 - 0.9 = -0.6 \, \text{kg m/s} \]

2. **Calculate the final velocity after the collision using conservation of momentum:**
- Let the final velocity of the combined mass be \( v \):
\[ (m + M) v = p_{\text{initial}} \implies (1 + 3)v = -0.6 \implies 4v = -0.6 \implies v = -0.15 \, \text{m/s} \]

3. **Calculate the new amplitude using conservation of energy:**
- Initial kinetic energy of the system:
\[ KE_{\text{initial}} = \frac{1}{2} m v_m^2 + \frac{1}{2} M v_M^2 = \frac{1}{2} \cdot 1 \cdot (0.3)^2 + \frac{1}{2} \cdot 3 \cdot (0.3)^2 \]
\[ = \frac{1}{2} \cdot 1 \cdot 0.09 + \frac{1}{2} \cdot 3 \cdot 0.09 = 0.045 + 0.135 = 0.18 \, \text{J} \]
- The spring constant \( k \) can be calculated from the SHM equation:
\[ \omega = 10 \, \text{rad/s} \implies k = m \omega^2 = 1 \cdot 10^2 = 100 \, \text{N/m} \]
- The potential energy in the spring at maximum amplitude \( A \):
\[ KE_{\text{final}} = \frac{1}{2} k A^2 \]
Setting \( KE_{\text{initial}} = KE_{\text{final}} \):
\[ 0.18 = \frac{1}{2} \cdot 100 \cdot A^2 \implies 0.18 = 50 A^2 \implies A^2 = \frac{0.18}{50} = 0.0036 \implies A = 0.06 \, \text{m} = 6 \, \text{cm} \]

### Part (b): New Equation for Position of the Combined Body

1. **Calculate the new angular frequency \( \omega' \):**
- The new mass is \( 4 \, \text{kg} \):
\[ \omega' = \sqrt{\frac{k}{m + M}} = \sqrt{\frac{100}{4}} = \sqrt{25} = 5 \, \text{rad/s} \]

2. **Write the new equation of motion:**
- The new equation of position will be:
\[ x = 10 + 6 \sin(5t) \quad (\text{in cm}) \]

### Part (c): Loss of Energy During Collision

1. **Calculate the energy before and after the collision:**
- Initial total energy before collision:
\[ E_{\text{initial}} = KE_{\text{initial}} = 0.18 \, \text{J} \]
- Final total energy after collision (using new amplitude):
\[ E_{\text{final}} = \frac{1}{2} k A^2 = \frac{1}{2} \cdot 100 \cdot (0.06)^2 = 0.18 \, \text{J} \]

2. **Calculate the loss of energy:**
- Since both energies are equal, the loss of energy during the collision is:
\[ \text{Loss of Energy} = E_{\text{initial}} - E_{\text{final}} = 0.18 - 0.18 = 0 \, \text{J} \]

### Summary of Results:
- (a) New amplitude of oscillations: \( 6 \, \text{cm} \)
- (b) New equation for position: \( x = 10 + 6 \sin(5t) \)
- (c) Loss of energy during collision: \( 0 \, \text{J} \)