## Problem Statement (Translation) **3.6** A delicate instrument must be mounted using an **isolation spring**, separating it from the laboratory floor, which is known to vibrate vertically with a harmonic motion of amplitude **0.25 cm** at **10 cycles per second (cps)**. If the instrument weighs **50 kp** (kiloponds), determine the required stiffness of the isolation spring to reduce the vertical amplitude of the instrument’s motion to **0.025 cm**. Ignore damping in the system. --- ## Step-by-Step Solution ### **1. Given Data** - **Amplitude of floor vibration**: \( Y = 0.25\, \text{cm} \) - **Frequency of vibration**: \( f = 10\, \text{Hz} \) - **Amplitude of instrument motion (desired)**: \( X = 0.025\, \text{cm} \) - **Weight of instrument**: \( W = 50\, \text{kp} \) - **Neglect damping** ### **2. Useful Concepts** - **Transmissibility (T)**: The ratio of the amplitude of the instrument to the amplitude of the base (floor): \[ T = \frac{X}{Y} = \frac{1}{\sqrt{\left[1 - \left(\frac{\omega}{\omega_n}\right)^2\right]^2}} \] (for undamped systems, since damping is neglected). - **Natural Frequency (\( \omega_n \))**: \[ \omega_n = \sqrt{\frac{k}{m}} \] where \( k \) is the spring stiffness, \( m \) is the mass. - **Weight and Mass**: \[ W = m g \implies m = \frac{W}{g} \] Take \( g = 9.81\, \text{m/s}^2 \). ### **3. Convert Units** - \( Y = 0.25\, \text{cm} = 0.0025\, \text{m} \) - \( X = 0.025\, \text{cm} = 0.00025\, \text{m} \) - \( W = 50\, \text{kp} = 50 \times 9.81\, \text{N} = 490.5\, \text{N} \) - \( f = 10\, \text{Hz} \implies \omega = 2\pi f = 62.83\, \text{rad/s} \) ### **4. Find Transmissibility** \[ T = \frac{X}{Y} = \frac{0.00025}{0.0025} = 0.1 \] ### **5. Rearranging the Transmissibility Equation** For undamped systems: \[ T = \frac{1}{|1 - r^2|} \] where \( r = \frac{\omega}{\omega_n} \) So, \[ 0.1 = \frac{1}{|1 - r^2|} \implies |1 - r^2| = 10 \implies r^2 = 1 \pm 10 \] Choose the plus sign (since frequency ratio should be greater than 1 for attenuation): \[ r^2 = 11 \implies r = \sqrt{11} \implies \frac{\omega}{\omega_n} = \sqrt{11} \implies \omega_n = \frac{\omega}{\sqrt{11}} \] ### **6. Calculate the Required Stiffness** \[ \omega_n = \sqrt{\frac{k}{m}} \implies \sqrt{\frac{k}{m}} = \frac{62.83}{\sqrt{11}} \] \[ k = m \left( \frac{62.83}{\sqrt{11}} \right)^2 \] First, calculate mass \( m \): \[ m = \frac{490.5}{9.81} = 50\, \text{kg} \] Now plug in the values: \[ k = 50 \left( \frac{62.83}{3.317} \right)^2 = 50 \times (18.94)^2 \approx 50 \times 359 \approx 17950\, \text{N/m} \] --- ## **Final Answer** ### **The required stiffness of the isolation spring is approximately:** \[ \boxed{17,\!950\, \text{N/m}} \] --- ### **Summary of Steps** 1. Found the transmissibility ratio. 2. Used the undamped transmissibility formula to solve for the frequency ratio. 3. Related natural frequency to stiffness. 4. Substituted all values to solve for spring stiffness. **Let me know if you need more explanation on any specific step!**