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 _________.
1.58 N/mm
Understanding the properties of vibrating systems is crucial in mechanical engineering. This problem involves a suspended mass attached to a spring, forming a single-degree-of-freedom system. Even though the system is described as damped, when damping parameters (like damping ratio) are not provided and the spring stiffness is required, the given oscillation data is typically used to determine the undamped natural frequency for calculating the spring's stiffness.
Oscillation Frequency Calculation
First, we need to calculate the frequency of oscillations from the given data. The suspended mass completes 40 oscillations in 20 seconds.
- Total number of oscillations (\(n\)) = 40
- Total time taken (\(t\)) = 20 seconds
The frequency (\(f\)) of oscillation is the number of oscillations per unit time:
\[ f = \frac{\text{Number of oscillations}}{\text{Time taken}} \]
Substituting the given values into the frequency formula:
\[ f = \frac{40 \text{ oscillations}}{20 \text{ seconds}} \]
\[ f = 2 \text{ oscillations/second} \]
Or, in standard units:
\[ f = 2 \text{ Hz} \]
Angular Natural Frequency
Next, we convert the frequency in Hertz (Hz) to angular natural frequency (\(\omega_n\)) in radians per second (rad/s). The relationship between linear frequency and angular frequency is:
\[ \omega_n = 2\pi f \]
Substituting the calculated frequency into the formula:
\[ \omega_n = 2\pi (2 \text{ Hz}) \]
\[ \omega_n = 4\pi \text{ rad/s} \]
Using the approximate value of \(\pi \approx 3.14159\):
\[ \omega_n = 4 \times 3.14159 \text{ rad/s} \]
\[ \omega_n \approx 12.56636 \text{ rad/s} \]
Stiffness Calculation for the Spring
For a single-degree-of-freedom undamped vibrating system, the angular natural frequency (\(\omega_n\)) is related to the mass (\(m\)) and spring stiffness (\(k\)) by the formula:
\[ \omega_n = \sqrt{\frac{k}{m}} \]
We are given the suspended mass (\(m\)) = 10 kg.
To find the stiffness (\(k\)) of the spring, we can rearrange the formula:
\[ \omega_n^2 = \frac{k}{m} \]
\[ k = m \omega_n^2 \]
Substituting the values of mass and angular natural frequency into the stiffness formula:
\[ k = 10 \text{ kg} \times (12.56636 \text{ rad/s})^2 \]
\[ k = 10 \text{ kg} \times 157.9136 \text{ (N/m)} \]
\[ k \approx 1579.136 \text{ N/m} \]
Converting Stiffness to N/mm
The options provided are given in N/mm, so we need to convert the calculated stiffness from N/m to N/mm. We know that 1 meter is equal to 1000 millimeters.
\[ k_{\text{N/mm}} = \frac{1579.136 \text{ N}}{1 \text{ m}} \times \frac{1 \text{ m}}{1000 \text{ mm}} \]
\[ k_{\text{N/mm}} = \frac{1579.136}{1000} \text{ N/mm} \]
\[ k_{\text{N/mm}} \approx 1.579136 \text{ N/mm} \]
Rounding the result to two decimal places, the stiffness of the spring is approximately 1.58 N/mm.
Summary of Key Concepts
This problem illustrates the fundamental principles of mechanical vibrations. We used the given oscillation data (number of oscillations and time) to determine the system's natural frequency. This natural frequency, combined with the suspended mass, allowed us to calculate the unknown spring stiffness. Although the system is described as damped, in the absence of damping coefficient information, the observed oscillation frequency is often treated as the undamped natural frequency for determining basic system parameters like spring stiffness. This is a common simplification in introductory vibration analysis to find inherent properties of the system.
| Parameter | Value | Unit |
|---|---|---|
| Suspended Mass (\(m\)) | 10 | kg |
| Number of Oscillations (\(n\)) | 40 | - |
| Time (\(t\)) | 20 | seconds |
| Frequency (\(f\)) | 2 | Hz |
| Angular Natural Frequency (\(\omega_n\)) | \(4\pi\) or 12.566 | rad/s |
| Stiffness (\(k\)) in N/m | 1579.136 | N/m |
| Stiffness (\(k\)) in N/mm | 1.579136 \(\approx\) 1.58 | N/mm |
The calculated value for the stiffness of the spring is approximately 1.58 N/mm, which matches option 4.
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