A mass of 2 kg is hung from the ceiling by a helical spring. When hung, the spring suffers an extension of 100 mm. If the mass is slightly displaced downward and released, it will oscillate at a frequency of (acceleration due gravity at the location is 10 m/s2)
1.6 Hz
This problem involves a mass attached to a helical spring, forming a simple harmonic oscillator. When a mass is suspended from a spring and allowed to hang in equilibrium, the spring stretches due to the weight of the mass. This extension allows us to determine the spring constant. When the mass is then slightly displaced from its equilibrium position and released, it oscillates up and down. We need to find the frequency of this oscillation.
When the mass is hanging at rest, it is in equilibrium. The forces acting on the mass are the downward gravitational force (weight) and the upward force exerted by the spring (spring force). According to Hooke's Law, the spring force is proportional to the extension of the spring.
At equilibrium, the weight of the mass is balanced by the spring force:
Weight = Spring Force
The weight is given by \(W = m \times g\).
The spring force is given by \(F_s = k \times \Delta L\), where \(k\) is the spring constant.
So, we have the equation:
\(m \times g = k \times \Delta L\)
We can rearrange this equation to solve for the spring constant \(k\):
\(k = \frac{m \times g}{\Delta L}\)
Substitute the given values:
\(k = \frac{2 \, \text{kg} \times 10 \, \text{m/s}^2}{0.1 \, \text{m}}\)
\(k = \frac{20 \, \text{N}}{0.1 \, \text{m}}\)
\(k = 200 \, \text{N/m}\)
The spring constant is \(200 \, \text{N/m}\).
For a mass-spring system undergoing simple harmonic motion, the angular frequency (\(\omega\)) is given by the formula:
\(\omega = \sqrt{\frac{k}{m}}\)
Where:
Substitute the calculated value of \(k\) and the given value of \(m\):
\(\omega = \sqrt{\frac{200 \, \text{N/m}}{2 \, \text{kg}}}\)
\(\omega = \sqrt{100} \, \text{rad/s}\)
\(\omega = 10 \, \text{rad/s}\)
The angular frequency is \(10 \, \text{rad/s}\). The frequency (\(f\)) of oscillation in Hertz (Hz) is related to the angular frequency by the equation:
\(\omega = 2 \pi f\)
Rearranging to solve for \(f\):
\(f = \frac{\omega}{2 \pi}\)
Substitute the value of \(\omega\):
\(f = \frac{10 \, \text{rad/s}}{2 \pi}\)
\(f = \frac{5}{\pi} \, \text{Hz}\)
To find the numerical value, we use an approximate value for \(\pi\), such as \(3.14159\):
\(f \approx \frac{5}{3.14159} \, \text{Hz}\)
\(f \approx 1.5915 \, \text{Hz}\)
Rounding to one decimal place, we get approximately \(1.6 \, \text{Hz}\).
Let's compare our calculated frequency with the given options:
| Option | Value | Comparison |
|---|---|---|
| 1 | 1.6 Hz | Matches our calculated value (approximately) |
| 2 | \(\sqrt{50}\) Hz \(\approx 7.07\) Hz | Does not match |
| 3 | 10 Hz | Does not match (This is the angular frequency in rad/s, not frequency in Hz) |
| 4 | 50 Hz | Does not match |
The calculated frequency of approximately 1.6 Hz matches Option 1.
| Concept | Formula | Description |
|---|---|---|
| Hooke's Law (Magnitude) | \(F_s = kx\) | Spring force is proportional to extension/compression |
| Spring Constant | \(k = \frac{F}{\Delta L}\) (from equilibrium) | Force per unit extension/compression |
| Angular Frequency (Mass-Spring) | \(\omega = \sqrt{\frac{k}{m}}\) | Rate of oscillation in radians per second |
| Frequency | \(f = \frac{\omega}{2\pi}\) | Number of oscillations per second (Hz) |
| Period | \(T = \frac{2\pi}{\omega} = \frac{1}{f}\) | Time taken for one complete oscillation |
Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement from the equilibrium position and acts towards the equilibrium. The mass-spring system, when friction is negligible, is a classic example of SHM.
In SHM, the frequency and period depend on the physical properties of the system (mass and spring constant) but are independent of the amplitude of oscillation, provided the oscillations are small enough for Hooke's Law to apply.
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