A spring-mass system has a natural frequency of 10 [Hz] When the spring constant is reduced by 800 N/m the frequency is altered by 45%. Find the mass and spring constant of the original system.
In the description of the example 3 we know that the spring system has a natural frequency of 10 [Hz].
If the spring constant is reduced by 800 N/m the frequency is altered 45%. So the old spring constant will be labeled as \(k_{old}\) while the new spring constant will be labeled as \(k_{new}\). To obtain the new value of spring constant we must subtract 800 from the old one, so we can write: \(k_{new} = k_{old} - 800\).
The natural frequncey is altered by 45% which means that the new natural frequency \(\omega_{n-new}\) is \((1-0.45)*\omega_{n-old}\) of the old one. To summarize the given data can be written as: $$f = 10 [\mathrm{Hz}]$$ $$ k_{new} = k_{old}-800$$ $$ \omega_{n-new} = (1-0.45)\omega_{n-old} $$ In this example we will need two formulas i.e. the formulas for calculating the natural angular frequency from natural frequency and the term for determine the natural angular frequency from ration of spring constant and mass. $$\omega_n = 2\pi f$$ $$\omega_n =\sqrt{\frac{k}{m}}$$ First step is to calculate the natural angular frequency from the natural frequency. $$\omega_n = 2*\pi*f = 2 * \pi * 10 = 62.832 [\mathrm{\frac{rad}{s}}]$$ $$\omega_{n-old} = \sqrt{\frac{k}{m}}$$ From the previous expression we can obtain the expression for determining the mass: $$ \sqrt{m} = \frac{\sqrt{k}}{62.832}$$ As stated previously the new natural angular frequency is the 0.55 of the old natural angular frequency: $$ \omega_{n-new} = 0.55 \omega_{n-old} = 34.5576 \left[\mathrm{\frac{rad}{s}}\right] $$ Now using the expression for natural angular frequency that requires spring constant and mass written in the following form: $$\omega_{n1} = \sqrt{\frac{k_{new}}{m_{new}}} = \sqrt{\frac{k-800}{m}}\\ $$ we can determine the spring constant of the new system. We akso know the the new spring cosntant is reduced by 800 N/m from the old one. $$\sqrt{\frac{k-800}{k}} \cdot 62.836 = 34.5576$$ $$\sqrt{\frac{k-800}{k}} = 0.55$$ $$\frac{k-800}{k} = 0.55^2$$ $$ k = 1146.9534 \left[\mathrm{\frac{N}{m}}\right]$$ After we have determine the spirng constant we can determine the mass of the spring mass system. $$ m = \frac{k}{52.832^2} = 0.2905[\mathrm{kg}] $$
Three libraries will be required i.e. numpy, sympy, and matplotlib. The sympy library will be necessary for solving the equation.
Solution to FVSDOF-SYS Example 3
In the description of the example 3 we know that the spring system has a natural frequency of 10 [Hz].
If the spring constant is reduced by 800 N/m the frequency is altered 45%. So the old spring constant will be labeled as \(k_{old}\) while the new spring constant will be labeled as \(k_{new}\). To obtain the new value of spring constant we must subtract 800 from the old one, so we can write: \(k_{new} = k_{old} - 800\).The natural frequncey is altered by 45% which means that the new natural frequency \(\omega_{n-new}\) is \((1-0.45)*\omega_{n-old}\) of the old one. To summarize the given data can be written as: $$f = 10 [\mathrm{Hz}]$$ $$ k_{new} = k_{old}-800$$ $$ \omega_{n-new} = (1-0.45)\omega_{n-old} $$ In this example we will need two formulas i.e. the formulas for calculating the natural angular frequency from natural frequency and the term for determine the natural angular frequency from ration of spring constant and mass. $$\omega_n = 2\pi f$$ $$\omega_n =\sqrt{\frac{k}{m}}$$ First step is to calculate the natural angular frequency from the natural frequency. $$\omega_n = 2*\pi*f = 2 * \pi * 10 = 62.832 [\mathrm{\frac{rad}{s}}]$$ $$\omega_{n-old} = \sqrt{\frac{k}{m}}$$ From the previous expression we can obtain the expression for determining the mass: $$ \sqrt{m} = \frac{\sqrt{k}}{62.832}$$ As stated previously the new natural angular frequency is the 0.55 of the old natural angular frequency: $$ \omega_{n-new} = 0.55 \omega_{n-old} = 34.5576 \left[\mathrm{\frac{rad}{s}}\right] $$ Now using the expression for natural angular frequency that requires spring constant and mass written in the following form: $$\omega_{n1} = \sqrt{\frac{k_{new}}{m_{new}}} = \sqrt{\frac{k-800}{m}}\\ $$ we can determine the spring constant of the new system. We akso know the the new spring cosntant is reduced by 800 N/m from the old one. $$\sqrt{\frac{k-800}{k}} \cdot 62.836 = 34.5576$$ $$\sqrt{\frac{k-800}{k}} = 0.55$$ $$\frac{k-800}{k} = 0.55^2$$ $$ k = 1146.9534 \left[\mathrm{\frac{N}{m}}\right]$$ After we have determine the spirng constant we can determine the mass of the spring mass system. $$ m = \frac{k}{52.832^2} = 0.2905[\mathrm{kg}] $$
Solution in Python
Three libraries will be required i.e. numpy, sympy, and matplotlib. The sympy library will be necessary for solving the equation.
import numpy as npNow we will define the natural frequency and calculate the natural angular frequency.
import sympy as sp
import matplotlib.pyplot as plt
f = 10When you execute the previous code the output should be:
omega_n-old = 2*np.pi*10
print("omega_n-old = {}".format(omega_n-old))
omega_n-old = 62.83185307179586The new natural angular frequency is the 45% lower than the old natural angular frequency.
omega_n-new = (1-0.45)*omega_n-oldThe final step is to solve the equation \(\sqrt{\frac{k-800}{k}} * \omega_{n-old} - \omega_{n-new}\) using sympy library. To do that we will need to define the symbol 'k', define the equation and then solve the equation.
k = sp.symbols('k')
equation = sp.sqrt((k-800)/k)*62.836-34.5576
solutions = sp.solve(equation, k)
print("k = {}".format(solutions))
When previous code is executed the following output is obtained:
k = 1146.9534