Question

Object A is four times heavier than object B. Each object is attached to a spring, and the springs have equal spring constants. The two objects are then pulled from their equilibrium positions and released from rest. What is the ratio of the periods of the two oscillators if the amplitude of \(\mathrm{A}\) is half that of \(\mathrm{B}\) ? a) \(T_{\mathrm{A}}: T_{\mathrm{B}}=1: 4\) b) \(T_{\mathrm{A}}: T_{\mathrm{B}}=4: 1\) c) \(T_{\mathrm{A}}: T_{\mathrm{B}}=2: 1\) d) \(T_{\mathrm{A}}: T_{\mathrm{B}}=1: 2\)

Short answer

Given: - Object A is 4 times heavier than Object B. - The amplitude of Object A is half that of Object B. Answer: The ratio of the periods of the two oscillators is \(T_{\mathrm{A}}: T_{\mathrm{B}} = 2:1\).

Step-by-step solution

Find the formula for the periods of the two oscillators

To find the ratio of the periods of the two oscillators (\(T_{\mathrm{A}}\) and \(T_{\mathrm{B}}\)), we need to find the periods of Object A and Object B using the period formula: \(T = 2\pi\sqrt{\frac{m}{k}}\).

Express the mass of Object A in terms of Object B's mass

We are given that Object A is four times heavier than Object B. Let the mass of Object B be \(m_\mathrm{B}\), then the mass of Object A will be \(4m_\mathrm{B}\).

Calculate the periods for both objects

Using the formula \(T = 2\pi\sqrt{\frac{m}{k}}\), we can calculate the periods for Object A and Object B as follows: The period of Object A, \(T_\mathrm{A} = 2\pi\sqrt{\frac{4m_\mathrm{B}}{k}}\) The period of Object B, \(T_\mathrm{B} = 2\pi\sqrt{\frac{m_\mathrm{B}}{k}}\)

Calculate the ratio of the periods \(T_{\mathrm{A}}: T_{\mathrm{B}}\)

Now we will calculate the ratio of the periods \(T_{\mathrm{A}}: T_{\mathrm{B}}\). \(\frac{T_{\mathrm{A}}}{T_{\mathrm{B}}} = \frac{2 \pi \sqrt{\frac{4m_{\mathrm{B}}}{k}}}{2\pi \sqrt{\frac{m_{\mathrm{B}}}{k}}}\) Note that \(2\pi\) and \(k\) can be cancelled from the equation. \(\frac{T_{\mathrm{A}}}{T_{\mathrm{B}}} = \frac{\sqrt{\frac{4m_{\mathrm{B}}}{m_{\mathrm{B}}}}}{\sqrt{1}}\) The ratio simplifies to: \(\frac{T_{\mathrm{A}}}{T_{\mathrm{B}}} = \sqrt{4}\)

Simplify and find the answer

Taking the square root of 4, we find the ratio: \(\frac{T_{\mathrm{A}}}{T_{\mathrm{B}}} = 2\) So, the ratio of the periods of the two oscillators is \(T_{\mathrm{A}}: T_{\mathrm{B}} = 2:1\), which corresponds to answer option c).

Key concepts

Harmonic Oscillator

A harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force proportional to the displacement. This principle is found in various physical systems, but we often visualize it as a mass attached to a spring. When the mass is pulled and released, it oscillates around the equilibrium position.

One of the keys to understanding a harmonic oscillator is acknowledging that regardless of the amplitude (how far you pull or push the object from its rest position), the period (the time it takes to complete one full cycle of motion) remains constant, assuming there's no external resistance. This constant period is a signature characteristic of simple harmonic motion, which we'll explore further. The formula used in the original exercise, which is derived from this principle, is instrumental for calculating the period.

Spring Constant

The spring constant, denoted as 'k' in physics, is a measure of a spring's stiffness. It plays a critical role in determining the behavior of a harmonic oscillator. In the given exercise, the spring constant 'k' is shared between Object A and Object B. The constant nature of 'k' for both springs means that the spring's resistance to stretching or compressing is the same.

The formula for the period of an oscillator, \(T = 2\pi\sqrt{\frac{m}{k}}\), is directly influenced by the spring constant. A higher value of 'k' would result in a shorter period, indicating a stiffer spring that oscillates more quickly. Conversely, a lower 'k' value implies a spring is less stiff and would have a longer period for its oscillations. Since 'k' is the same for both springs in the problem, the difference in periods for objects A and B is solely due to their mass difference.

Simple Harmonic Motion

Simple Harmonic Motion (SHM) is the type of periodic motion that the harmonic oscillator undergoes. SHM can be defined as the motion of an object when the force acting on it is directly proportional to its displacement from the equilibrium position and is always directed towards that position.

In the context of our exercise, both Object A and Object B exhibit simple harmonic motion when they are attached to identical springs and displaced from the equilibrium position. The period of the simple harmonic motion is unaffected by amplitude; that explains why the amplitude of the objects does not factor into our calculations for the period. This principle is crucial in understanding why, although the masses of the objects differ, it is this difference, and not their respective amplitudes, that affects the ratio of their oscillation periods. The elegance of SHM allows for the clear-cut solution found in the exercise, where the mass relationship between the objects directly yields the period ratio.