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Question

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)

The correct answer is

1.6 Hz

Understanding Mass-Spring Oscillation

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.

Determining the Spring Constant (\(k\))

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.

  • Mass \(m = 2 \, \text{kg}\)
  • Extension \(\Delta L = 100 \, \text{mm} = 0.1 \, \text{m}\)
  • Acceleration due to gravity \(g = 10 \, \text{m/s}^2\)

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 \(200 \, \text{N/m}\).

Calculating the Oscillation Frequency

For a mass-spring system undergoing simple harmonic motion, the angular frequency (\(\omega\)) is given by the formula:

\(\omega = \sqrt{\frac{k}{m}}\)

Where:

  • \(k\) is the spring constant
  • \(m\) is the mass

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 \(1.6 \, \text{Hz}\).

Comparing with Options

Let's compare our calculated frequency with the given options:

OptionValueComparison
11.6 HzMatches our calculated value (approximately)
2\(\sqrt{50}\) Hz \(\approx 7.07\) HzDoes not match
310 HzDoes not match (This is the angular frequency in rad/s, not frequency in Hz)
450 HzDoes not match

The calculated frequency of approximately 1.6 Hz matches Option 1.

Revision Table: Key Formulas

ConceptFormulaDescription
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

Additional Information on Simple Harmonic Motion (SHM)

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.

  • Equilibrium Position: The position where the net force on the object is zero. For a vertical spring, this is where the spring force balances gravity and any other applied forces.
  • Restoring Force: The force that always tries to bring the object back to its equilibrium position. In a spring, this is the spring force \(F_s = -kx\) (the negative sign indicates it's a restoring force, opposite to displacement \(x\)).
  • Amplitude: The maximum displacement from the equilibrium position.
  • Period (\(T\)): The time it takes for one complete cycle of oscillation.
  • Frequency (\(f\)): The number of cycles per unit time, usually measured in Hertz (Hz), where \(1 \, \text{Hz} = 1\) cycle per second.
  • Angular Frequency (\(\omega\)): Related to frequency by \(\omega = 2\pi f\). It represents the rate of oscillation in radians per second.

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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Important Questions from Simple Mass System

  1. 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?

  2. 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:

  3. If mass M oscillates on a spring having mass m and stiffness k, then the natural frequency of the system is

  4. 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

  5. Which of the following statements is false with respect to a simple pendulum?

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