In a single-degree damped vibrating system, a suspended mass of 5 kg makes 21 oscillations in 11 seconds. The stiffness of the spring will be
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 (fn): The frequency at which the system would oscillate without damping, measured in Hertz (Hz).
- Natural Angular Frequency (ωn): 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 (fn)
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 (ωn)
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.
Important Questions from Damped Free Vibration
Which condition is suitable for indicating instruments in order to get the best results?
When there is reduction in amplitude over every cycle of vibration, then the body is said to have
Which of the following statements are TRUE for damped vibrations?
P. For a system having critical damping, the value of the damping ratio is unity and the system does not undergo a vibratory motion.
Q. Logarithmic decrement method is used to determine the amount of damping in a physical system.
R. In case of damping due to dry friction between moving surfaces resisting force of constant magnitude acts opposite to the relative motion.
S. For the case of viscous damping, drag force is directly proportional to the square of relative velocity.
The critical damping is said to occur when the frequency of damped vibration is
A suspended mass of 10 kg completes 40 oscillations in 20 seconds in a single-degree damped vibrating system. The stiffness of the spring is approximately _________.
