A horizontal spring-block system of mass 1kg executes SHM of amplitude 10 cm. When the block is passing through its equilibrium position another mass of 1kg is put on it and the two move together.

Options

A. amplitude will remain unchanged
B. amplitude will become `5 sqrt2` cm
C. the frequency of oscillations will remain same
D. the frequency of oscillations will decrease.

Answer

To solve the problem, we need to analyze the effects of adding an additional mass to a spring-block system that is executing simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- The initial mass \( m_1 = 1 \, \text{kg} \).
- The amplitude \( A = 10 \, \text{cm} = 0.1 \, \text{m} \).
- The system is executing SHM.

2. **Determine the Time Period Before Adding the Mass:**
- The time period \( T_1 \) of the system is given by:
\[ T_1 = 2\pi \sqrt{\frac{m_1}{k}} \]
- Here, \( k \) is the spring constant.

3. **Adding the Second Mass:**
- When the second mass \( m_2 = 1 \, \text{kg} \) is added, the total mass becomes:
\[ m_{\text{total}} = m_1 + m_2 = 1 \, \text{kg} + 1 \, \text{kg} = 2 \, \text{kg} \]

4. **Determine the Time Period After Adding the Mass:**
- The new time period \( T_2 \) is:
\[ T_2 = 2\pi \sqrt{\frac{m_{\text{total}}}{k}} = 2\pi \sqrt{\frac{2 \, \text{kg}}{k}} \]

5. **Determine the Frequencies:**
- The frequency \( f \) is the reciprocal of the time period:
\[ f_1 = \frac{1}{T_1} = \frac{1}{2\pi} \sqrt{\frac{k}{m_1}} \]
\[ f_2 = \frac{1}{T_2} = \frac{1}{2\pi} \sqrt{\frac{k}{m_{\text{total}}}} = \frac{1}{2\pi} \sqrt{\frac{k}{2 \, \text{kg}}} \]

6. **Compare the Frequencies:**
- Since \( m_{\text{total}} = 2 \, m_1 \), we can see that:
\[ f_2 = \frac{1}{\sqrt{2}} f_1 \]
- This indicates that the frequency decreases when the mass is doubled.

7. **Determine the Amplitude After Adding the Mass:**
- The kinetic energy at the equilibrium position before adding the mass is:
\[ KE = \frac{1}{2} k A^2 \]
- After adding the mass, the total energy remains the same, but the kinetic energy will be shared between the two masses:
\[ KE' = \frac{1}{2} (m_1 + m_2) v'^2 \]
- Using conservation of momentum, the velocity after adding the mass becomes:
\[ v' = \frac{m_1 v}{m_1 + m_2} = \frac{v}{2} \]
- The new amplitude \( A' \) can be found using the relationship between the velocities and amplitudes:
\[ A' = \frac{A}{\sqrt{2}} = \frac{10 \, \text{cm}}{\sqrt{2}} = 5\sqrt{2} \, \text{cm} \]

### Final Answers:
- **Amplitude remains unchanged:** No.
- **Amplitude is \( 5\sqrt{2} \, \text{cm} \):** Yes.
- **Frequency of oscillation remains the same:** No, it decreases.