FVSDOF-SYS Example 4 - Spring Fixed on Both Ends

The srping fixed at one and loaded on the other will produce the elongation of 10 mm if its loaded with 100 N of force. The ends of the spring are rigidly fixed, and one end and vertically above the other. A mass of 10kg is attached at the middle point of its length.Determine the time taken to complete one vibration cycle.

Solution to the FVDOF-SYS Example 4

The goal of this example is to calculate the time taken to complete one vibration cycle. To calculate the vibration cycle we need to determine the angular natural frequency \(\omega_n\) and the spring constant. First we need to determine the spring constant. We know that if the spring is loaded with 100 N the spring will be elongated by 10 mm. From this data the spring constant can be determined: $$k = \frac{100}{\frac{10}{1000}} = 10000\left[\frac{\mathrm{N}}{\mathrm{m}}\right]$$ After determining the spring constant we can easily calculate the natural angular frequency : $$ \omega_n = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{4k}{m}}$$ $$ \omega_n = \sqrt{\frac{4\cdot 10^4}{10}} = 63.2456 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]$$ $$ \tau_n = \frac{2\pi}{\omega_n} = \frac{6.28}{63.2456} = 0.0993 [\mathrm{s}]$$ Import numpy and matplotlib libraries

import numpy as np

Calculate the spring constant, however, convert the elongation from millimeters to meters and define the force.

x = 10/1000 = 0.001
F = 100
k = F/x
print("F = {}".format(F))

When you execute the previous code the following solution is obtained:

F = 10000

The natural angular frequency of the system:

omega_n = np.sqrt(4*k/10)
print("omega_n = {}".format(omega_n))

The output of the previous code so far:

F = 10000
omega_n = 63.245553203367585

The oscillation period is calculated as:

tau_n = (2*np.pi)/omega_n
print("tau_n = {}".format(tau_n))
       

The output of the previous block of code is given below:

tau_n = 0.09934