Two springs of stiffness 100 N/m each are connected in series and support a mass of 2 kg. The natural frequency of the system will be
Understanding Springs in Series and Natural Frequency
When springs are connected in series, they behave differently compared to when they are connected in parallel. In a series connection, the total extension is the sum of the extensions of individual springs. This results in a system that is less stiff than the individual springs. To find the natural frequency of a mass-spring system, we first need to determine the equivalent stiffness of the spring combination.
Calculating Equivalent Stiffness for Springs in Series
For springs connected in series, the reciprocal of the equivalent stiffness (\(k_{eq}\)) is the sum of the reciprocals of the individual stiffnesses (\(k_1, k_2, \dots\)). The formula is:
\(\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} + \dots\)
In this problem, we have two springs, each with a stiffness of 100 N/m, connected in series. Let \(k_1 = 100\) N/m and \(k_2 = 100\) N/m.
Using the formula for series connection:
\(\frac{1}{k_{eq}} = \frac{1}{100} + \frac{1}{100}\)
\(\frac{1}{k_{eq}} = \frac{2}{100}\)
\(\frac{1}{k_{eq}} = \frac{1}{50}\)
Therefore, the equivalent stiffness is:
\(k_{eq} = 50\) N/m
Calculating the Natural Frequency
The natural frequency (\(f\)) of a mass-spring system is the frequency at which the system oscillates freely without any driving force or damping. The formula for natural frequency is:
\(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\)
where:
- \(f\) is the natural frequency in Hertz (Hz)
- \(k\) is the stiffness of the spring (or equivalent stiffness) in N/m
- \(m\) is the mass attached to the spring in kg
In this problem, we have:
- Equivalent stiffness \(k_{eq} = 50\) N/m
- Mass \(m = 2\) kg
Substituting these values into the natural frequency formula:
\(f = \frac{1}{2\pi}\sqrt{\frac{50}{2}}\)
\(f = \frac{1}{2\pi}\sqrt{25}\)
\(f = \frac{1}{2\pi} \times 5\)
\(f = \frac{5}{2\pi}\) Hz
Comparing with Options
Let's compare our calculated frequency \(\frac{5}{2\pi}\) Hz with the given options:
Option 1: \(\frac{{\pi}}{{2.5}}\) Hz
Option 2: \(\frac{{2.5}}{{\pi}}\) Hz
Option 3: \(\frac{{\sqrt{50}}}{{2\pi}}\) Hz = \(\frac{5\sqrt{2}}{2\pi}\) Hz
Option 4: \(\sqrt{50}\) Hz = \(5\sqrt{2}\) Hz
Let's check if our result \(\frac{5}{2\pi}\) matches option 2, which is \(\frac{2.5}{\pi}\):
\(\frac{5}{2\pi} = \frac{2 \times 2.5}{2\pi} = \frac{2.5}{\pi}\)
Yes, our calculated natural frequency matches option 2.
Summary of Calculation
| Parameter | Value |
|---|---|
| Stiffness of each spring (\(k_1, k_2\)) | 100 N/m |
| Connection type | Series |
| Equivalent stiffness (\(k_{eq}\)) | 50 N/m |
| Mass (\(m\)) | 2 kg |
| Natural frequency (\(f\)) | \(\frac{5}{2\pi}\) Hz or \(\frac{2.5}{\pi}\) Hz |
The natural frequency of the system is found by calculating the equivalent stiffness of the springs in series and then using the formula for the natural frequency of a mass-spring system. The equivalent stiffness of two 100 N/m springs in series is 50 N/m. With a mass of 2 kg, the natural frequency is \(\frac{5}{2\pi}\) Hz, which is equivalent to \(\frac{2.5}{\pi}\) Hz.
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