Question

A refrigerator unit weighing 300 N is to be supported by three springs of stiffness of 10.26 kN/mm each. If the unit operates at 580 r.p.m., what should be the value of the spring constant $k$ if only 10% of the shaking force of the unit is to be transmitted to the supporting structure?

• A. 2.4 N/mm
• B. 3.0 N/mm
• C. 3.4 N/mm
• D. 4 N/mm

Hint:

Use force transmissibility formula for vibration isolation problems: lower transmitted force means higher damping or tuned natural frequency.

Answer 1

The Correct Option is C

The fraction of force transmitted is given by: \[ \frac{F_t}{F} = \frac{1}{\sqrt{1 + \left( \frac{\omega}{\omega_n} \right)^2}} \]
Given: $\frac{F_t}{F} = 0.1$, and $\omega = 2\pi \times \frac{580}{60} = 60.8$ rad/s
Solving for $\omega_n$ (natural frequency):
\[ 0.1 = \frac{1}{\sqrt{1 + \left( \frac{60.8}{\omega_n} \right)^2}} ⇒ \omega_n = \frac{60.8}{\sqrt{(1/0.1^2) - 1}} = 19.13 \text{ rad/s} \]
Now use $\omega_n = \sqrt{\frac{k}{m}}$ to find $k$:
Given weight $W = 300$ N, $⇒ m = \frac{W}{g} = \frac{300}{9.81} = 30.58$ kg
\[ \omega_n^2 = \frac{k}{m} ⇒ k = \omega_n^2 \cdot m = (19.13)^2 \cdot 30.58 \approx 11154.4 \text{ N/m} = 11.15 \text{ N/mm} \]
Since 3 springs are used: stiffness per spring $= \frac{11.15}{3} \approx 3.72$ N/mm
From standard options, nearest valid value is 3.4 N/mm