A block with mass `M` attached to a horizontal spring with force constant `k` is moving with simple harmonic motion having amplitude `A_(1)`. At the instant when the block passes through its equilibrium position a lump of putty with mass `m` is dropped vertically on the block from a very small height and sticks to it.
(a) Find the new amplitude and period.
(b) Repeat part (a) for the case in which the putty is dropped on the block when it is at one end of its path.

Explanation

To solve the problem step by step, we will break it down into two parts as requested:

### Part (a): New Amplitude and Period when Putty is Dropped at Equilibrium Position

1. **Identify Initial Conditions**:
- Mass of the block = \( M \)
- Spring constant = \( k \)
- Initial amplitude = \( A_1 \)
- Initial period \( T_1 = 2\pi\sqrt{\frac{M}{k}} \)

2. **Determine the New Mass**:
- When the putty of mass \( m \) sticks to the block, the new mass \( M' = M + m \).

3. **Calculate the New Period**:
- The new period \( T_2 \) can be calculated using the formula:
\[ T_2 = 2\pi\sqrt{\frac{M'}{k}} = 2\pi\sqrt{\frac{M+m}{k}} \]

4. **Determine the New Amplitude**:
- At the equilibrium position, the velocity of the block is maximum. The maximum velocity \( v_{max} \) of the block before the putty sticks can be calculated as:
\[ v_{max} = A_1 \omega_1 = A_1 \cdot \sqrt{\frac{k}{M}} \]
- After the putty sticks, the new amplitude \( A_2 \) can be found using conservation of momentum. The momentum before the putty sticks is equal to the momentum after:
\[ M v_{max} = (M + m) v_{new} \]
- Rearranging gives:
\[ v_{new} = \frac{M v_{max}}{M + m} \]
- The new amplitude can be calculated using:
\[ A_2 = \frac{v_{new}}{\omega_2} = \frac{M v_{max}}{(M + m) \sqrt{\frac{k}{M + m}}} \]
- Substituting \( v_{max} \) into the equation gives:
\[ A_2 = \frac{M A_1 \sqrt{\frac{k}{M}}}{(M + m) \sqrt{\frac{k}{M + m}}} \]

### Part (b): New Amplitude and Period when Putty is Dropped at One End of the Path

1. **Identify Initial Conditions**:
- The initial conditions remain the same as in part (a).

2. **Determine the New Mass**:
- The new mass \( M' = M + m \) remains unchanged.

3. **Calculate the New Period**:
- The new period \( T_2 \) remains the same as in part (a):
\[ T_2 = 2\pi\sqrt{\frac{M+m}{k}} \]

4. **Determine the New Amplitude**:
- When the putty is dropped at one end of the path, the block is momentarily at rest. Thus, the new amplitude \( A_2 \) can be calculated as:
- The total energy in the system before the putty sticks is:
\[ E = \frac{1}{2} k A_1^2 \]
- After the putty sticks, the energy will be:
\[ E' = \frac{1}{2} (M + m) v_{new}^2 + \frac{1}{2} k A_2^2 \]
- Setting the energies equal gives:
\[ \frac{1}{2} k A_1^2 = \frac{1}{2} (M + m) v_{new}^2 + \frac{1}{2} k A_2^2 \]
- Since \( v_{new} = 0 \) at the end of the path, we can simplify to find \( A_2 \):
\[ k A_1^2 = k A_2^2 \implies A_2 = A_1 \sqrt{\frac{M}{M+m}} \]

### Summary of Results:
- **New Period**:
\[ T_2 = 2\pi\sqrt{\frac{M+m}{k}} \]
- **New Amplitude**:
- When dropped at equilibrium position:
\[ A_2 = A_1 \sqrt{\frac{M}{M+m}} \]
- When dropped at one end of the path:
\[ A_2 = A_1 \sqrt{\frac{M}{M+m}} \]