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.
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
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 npCalculate the spring constant, however, convert the elongation from millimeters to meters and define the force.
x = 10/1000 = 0.001When you execute the previous code the following solution is obtained:
F = 100
k = F/x
print("F = {}".format(F))
F = 10000The natural angular frequency of the system:
omega_n = np.sqrt(4*k/10)The output of the previous code so far:
print("omega_n = {}".format(omega_n))
F = 10000The oscillation period is calculated as:
omega_n = 63.245553203367585
tau_n = (2*np.pi)/omega_nThe output of the previous block of code is given below:
print("tau_n = {}".format(tau_n))
tau_n = 0.09934