Let's break down the problem step-by-step! --- ## **Step 1: Understand the Problem** - **Mass of putty, \( m_p \) = 2.9 kg** - **Mass of block, \( m_b \) = 31 kg** - **Spring constant for each spring, \( k = 740 \) N/m** - **Number of springs = 4** - **Putty is dropped from \( h = 2.2 \) m** - **You are to find the displacement \( x \) as a function of time after the putty sticks to the block.** --- ## **Step 2: Total Mass and Total Spring Constant** - **Total Mass, \( M = m_p + m_b = 2.9 + 31 = 33.9 \) kg** - **Total Spring Constant (since springs are in parallel):** \[ k_{total} = 4 \times 740 = 2960 \; \text{N/m} \] --- ## **Step 3: Initial Conditions** ### **A. Velocity of Putty Just Before Impact** Using energy conservation: \[ v = \sqrt{2gh} \] \[ v = \sqrt{2 \times 9.81 \times 2.2} = \sqrt{43.164} \approx 6.574 \; \text{m/s} \] ### **B. Initial Velocity of System** Just after the putty sticks, the velocity (by conservation of momentum): \[ m_p v_{putty} + m_b \cdot 0 = (m_p + m_b) v_{0} \] \[ v_{0} = \frac{m_p v_{putty}}{m_p + m_b} = \frac{2.9 \times 6.574}{33.9} \approx 0.562 \; \text{m/s} \] --- ## **Step 4: Differential Equation for SHM** For mass-spring system: \[ M\ddot{x} + k_{total} x = 0 \] \[ \ddot{x} + \omega^2 x = 0 \quad \text{where} \quad \omega = \sqrt{\frac{k_{total}}{M}} \] \[ \omega = \sqrt{\frac{2960}{33.9}} \approx \sqrt{87.314} \approx 9.35 \; \text{rad/s} \] --- ## **Step 5: General Solution** \[ x(t) = A \cos(\omega t) + B \sin(\omega t) \] #### **Initial Conditions:** - \( x(0) = 0 \) (measured from equilibrium) - \( \dot{x}(0) = v_0 = 0.562 \) m/s Plug into the general solution: - At \( t = 0 \): \( x(0) = A = 0 \implies A = 0 \) - \( \dot{x}(t) = B \omega \cos(\omega t) \) - At \( t = 0 \): \( \dot{x}(0) = B \omega = v_0 \implies B = \frac{v_0}{\omega} = \frac{0.562}{9.35} \approx 0.0601 \) So, \[ x(t) = 0.0601 \sin(9.35t) \] --- ## **Step 6: Plug in the Times (Convert to mm)** ### \( t = 0.044 \) s \[ x(0.044) = 0.0601 \sin(9.35 \times 0.044) \] \[ = 0.0601 \sin(0.4114) \] \[ = 0.0601 \times 0.3996 \approx 0.0240 \; \text{m} = 24.0 \; \text{mm} \] ### \( t = 0.260 \) s \[ x(0.260) = 0.0601 \sin(9.35 \times 0.260) \] \[ = 0.0601 \sin(2.431) \] \[ = 0.0601 \times 0.6542 \approx 0.0393 \; \text{m} = 39.3 \; \text{mm} \] ### \( t = 0.558 \) s \[ x(0.558) = 0.0601 \sin(9.35 \times 0.558) \] \[ = 0.0601 \sin(5.218) \] \[ = 0.0601 \times -0.8727 \approx -0.0524 \; \text{m} = -52.4 \; \text{mm} \] --- ## **Final Answers** | Time (s) | \( x \) (mm) | |----------|--------------| | 0.044 | 24.0 | | 0.260 | 39.3 | | 0.558 | -52.4 | --- ### **Summary** - **Displacement as a function of time:** \[ x(t) = 0.0601 \sin(9.35t) \quad \text{(in meters)} \] - Substitute values for \( t \) to get displacements in mm. Let me know if you need a detailed explanation of any step!