Calculating Spring Stiffness in a Damped System

This solution explains how to determine the stiffness of a spring in a single-degree damped vibrating system, given the mass and the number of oscillations over a specific time period.

Understanding the Concepts

• Mass (m): The suspended mass in the system, given as 5 kg.
• Oscillations: The complete back-and-forth movements of the mass.
• Time Period (T): The time taken for one complete oscillation.
• Natural Frequency (f_n): The frequency at which the system would oscillate without damping, measured in Hertz (Hz).
• Natural Angular Frequency (ω_n): The natural frequency expressed in radians per second (rad/s).
• Stiffness (k): The spring's resistance to deformation, measured in N/mm or N/m.

Step-by-Step Calculation

1. Determine the Time Period (T)

The time period is the total time divided by the number of oscillations.

Given:

• Number of oscillations = 21
• Total time = 11 seconds

Calculation:

$$ T = \frac{\text{Total time}}{\text{Number of oscillations}} = \frac{11 \text{ s}}{21 \text{ oscillations}} $$

$$ T = \frac{11}{21} \text{ s} $$

2. Calculate the Natural Frequency (f_n)

The natural frequency is the reciprocal of the time period.

Calculation:

$$ f_n = \frac{1}{T} = \frac{1}{\frac{11}{21} \text{ s}} = \frac{21}{11} \text{ Hz} $$

3. Calculate the Natural Angular Frequency (ω_n)

The natural angular frequency is related to the natural frequency by the formula $\( \omega_n = 2 \pi f_n \)$ .

Calculation:

$$ \omega_n = 2 \pi \times \frac{21}{11} \text{ rad/s} $$

4. Calculate the Spring Stiffness (k)

The relationship between natural angular frequency, mass, and stiffness is given by the formula $\( \omega_n = \sqrt{\frac{k}{m}} \)$ . Squaring both sides gives $\( \omega_n^2 = \frac{k}{m} \)$ .

Rearranging the formula to solve for stiffness (k):

$$ k = m \times \omega_n^2 $$

Substitute the known values:

Mass, $\( m = 5 \text{ kg} \)$

Natural angular frequency, $\( \omega_n = 2 \pi \frac{21}{11} \text{ rad/s} \)$

$$ k = 5 \text{ kg} \times \left( 2 \pi \frac{21}{11} \text{ rad/s} \right)^2 $$

$$ k = 5 \times (2 \pi)^2 \times \left(\frac{21}{11}\right)^2 \text{ N/m} $$

$$ k = 5 \times 4 \pi^2 \times \frac{441}{121} \text{ N/m} $$

Using the approximate value $\( \pi^2 \approx 9.8696 \)$ :

$$ k \approx 20 \times 9.8696 \times \frac{441}{121} \text{ N/m} $$

$$ k \approx 197.392 \times 3.6446 \text{ N/m} $$

$$ k \approx 719.42 \text{ N/m} $$

5. Convert Units from N/m to N/mm

The options are given in Newtons per millimeter (N/mm). Since 1 meter = 1000 millimeters, we need to divide the stiffness value in N/m by 1000.

$$ k_{\text{N/mm}} = \frac{k_{\text{N/m}}}{1000} $$

$$ k \approx \frac{719.42 \text{ N/m}}{1000} \approx 0.71942 \text{ N/mm} $$

Conclusion

The calculated stiffness is approximately 0.71942 N/mm, which is closest to 0.72 N/mm.