A simple spring mass vibrating system has a natural frequency of fn. If the spring stiffness is halved and mass is double, then the natural frequency will become
Natural Frequency of a Spring-Mass System
A simple spring-mass vibrating system has a natural frequency that depends directly on the spring's stiffness and the attached mass. To understand how changes in these parameters affect the system's vibration, we use a specific formula.
Formula for Natural Frequency
The natural frequency (\(f\)) of a simple spring-mass system is defined by the following equation:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
- Here, \(f\) represents the natural frequency, typically measured in Hertz (Hz).
- \(k\) is the spring stiffness, also known as the spring constant, measured in Newtons per meter (N/m).
- \(m\) is the mass attached to the spring, measured in kilograms (kg).
- \(\pi\) (pi) is a mathematical constant.
Initial Natural Frequency Condition
Let's consider the initial state of the system. The natural frequency is given as \(f_n\). We can denote the initial spring stiffness as \(k_1\) and the initial mass as \(m_1\).
So, we can write the initial natural frequency as:
\[ f_n = \frac{1}{2\pi} \sqrt{\frac{k_1}{m_1}} \quad \text{(Equation 1)} \]
Changes in Spring Stiffness and Mass
The problem states that the system undergoes two changes:
- The spring stiffness is halved. This means the new stiffness, \(k_2\), will be \(k_1\) divided by 2: \(k_2 = \frac{k_1}{2}\).
- The mass is doubled. This means the new mass, \(m_2\), will be \(m_1\) multiplied by 2: \(m_2 = 2m_1\).
Calculating the New Natural Frequency
Now, we will calculate the new natural frequency, let's call it \(f'\), by substituting the new values of stiffness and mass into the natural frequency formula:
\[ f' = \frac{1}{2\pi} \sqrt{\frac{k_2}{m_2}} \]
Substitute \(k_2 = \frac{k_1}{2}\) and \(m_2 = 2m_1\):
\[ f' = \frac{1}{2\pi} \sqrt{\frac{\frac{k_1}{2}}{2m_1}} \]
To simplify the expression inside the square root, we multiply the denominator by the denominator of the numerator:
\[ f' = \frac{1}{2\pi} \sqrt{\frac{k_1}{2 \times 2m_1}} \]
\[ f' = \frac{1}{2\pi} \sqrt{\frac{k_1}{4m_1}} \]
We can separate the constant factor \(\frac{1}{4}\) from the terms involving \(k_1\) and \(m_1\):
\[ f' = \frac{1}{2\pi} \sqrt{\frac{1}{4} \times \frac{k_1}{m_1}} \]
Using the property of square roots \(\sqrt{ab} = \sqrt{a} \sqrt{b}\):
\[ f' = \frac{1}{2\pi} \times \sqrt{\frac{1}{4}} \times \sqrt{\frac{k_1}{m_1}} \]
The square root of \(\frac{1}{4}\) is \(\frac{1}{2}\):
\[ f' = \frac{1}{2\pi} \times \frac{1}{2} \times \sqrt{\frac{k_1}{m_1}} \]
Rearrange the terms to group the original natural frequency expression:
\[ f' = \frac{1}{2} \left( \frac{1}{2\pi} \sqrt{\frac{k_1}{m_1}} \right) \]
From Equation 1, we know that \(\left( \frac{1}{2\pi} \sqrt{\frac{k_1}{m_1}} \right)\) is equal to \(f_n\). Substituting \(f_n\) into the equation for \(f'\):
\[ f' = \frac{1}{2} f_n \]
Therefore, the new natural frequency becomes half of the original natural frequency.
Summary of Frequency Change
Here's a summary of how the natural frequency changes with modifications to stiffness and mass:
| Parameter | Initial State | New State | Impact on Natural Frequency |
|---|---|---|---|
| Spring Stiffness (\(k\)) | \(k_1\) | \(k_1/2\) (Halved) | Since \(f \propto \sqrt{k}\), halving \(k\) contributes a factor of \(\sqrt{1/2}\). |
| Mass (\(m\)) | \(m_1\) | \(2m_1\) (Doubled) | Since \(f \propto 1/\sqrt{m}\), doubling \(m\) contributes a factor of \(1/\sqrt{2}\). |
| Combined Effect | \(f_n\) | \(f'\) | The combined effect is \(f' = (\sqrt{1/2} \times 1/\sqrt{2}) \times f_n = (1/\sqrt{4}) \times f_n = (1/2) \times f_n\). |
This detailed analysis shows that when the spring stiffness is halved and the mass is doubled, the natural frequency of the spring-mass vibrating system becomes \(\frac{f_n}{2}\).
Important Questions from Simple Mass System
A flexible rotor-shaft system comprises of a 10 kg rotor disc placed in the middle of a massless shaft of diameter 30 mm and length 500 mm between bearings (shaft is being taken mass-less as the equivalent mass of the shaft is included in the rotor mass) mounted at the ends. The bearings are assumed to simulate simply supported boundary conditions. The shaft is made of steel for which the value of E is 2.1 x 1011 Pa. What is the critical speed of rotation of the shaft?
Natural frequency (ωn) of a passenger car whose weight is w Newton and whose suspension has a combined stiffness of k N/mm is given by:
If mass M oscillates on a spring having mass m and stiffness k, then the natural frequency of the system is
Which of the following statements is false with respect to a simple pendulum?
The equation of motion for a spring-mass system excited by a harmonic force is
\(M\ddot x + kx = F\cos \left( {\omega t} \right),\)
Where M is the mass, K is the spring stiffness, F is the force amplitude and ω is the angular frequency of excitation. Resonance occurs when ω is equal to
