Question

A block having mass \(\mathrm{M}\) is placed on a horizontal frictionless surface. This mass is attached to one end of a spring having force constant \(\mathrm{k}\). The other end of the spring is attached to a rigid wall. This system consisting of spring and mass \(\mathrm{M}\) is executing SHM with amplitude \(\mathrm{A}\) and frequency \(\mathrm{f}\). When the block is passing through the mid-point of its path of motion, a body of mass \(\mathrm{m}\) is placed on top of it, as a result of which its amplitude and frequency changes to \(\mathrm{A}^{\prime}\) and \(\mathrm{f}\). The ratio of frequencies \((\mathrm{f} / \mathrm{f})=\ldots \ldots \ldots\) (A) \(\sqrt{\\{} \mathrm{M} /(\mathrm{m}+\mathrm{M})\\}\) (B) \(\sqrt{\\{\mathrm{m} /(\mathrm{m}+\mathrm{M})\\}}\) (C) \(\sqrt{\\{\mathrm{MA} / \mathrm{mA}}\\}\) (D) \(\sqrt{[}\\{(\mathrm{M}+\mathrm{m}) \mathrm{A}\\} / \mathrm{mA}]\)

Short answer

None of the given options are correct. The ratio of the initial frequency to the final frequency is: \(\frac{f}{f'} = \sqrt{1 + \frac{m}{M}}\)

Step-by-step solution

Frequency of Simple Harmonic Motion

The frequency of a mass-spring system undergoing simple harmonic motion is given by the following formula: \(f = \frac{1}{2 \pi} \sqrt{\frac{k}{M}}\) Initially, the frequency of the block with mass M is: \(f = \frac{1}{2 \pi} \sqrt{\frac{k}{M}}\)

Frequency of the Combined System

When mass m is added to the system, the total mass becomes (M + m), and the frequency of the combined system is given by: \(f' = \frac{1}{2 \pi} \sqrt{\frac{k}{M + m}}\)

Calculate the Ratio of Frequencies

Now, we want to find the ratio of the initial frequency (f) to the final frequency (f'): \(\frac{f}{f'} = \frac{\frac{1}{2 \pi} \sqrt{\frac{k}{M}}}{\frac{1}{2 \pi} \sqrt{\frac{k}{M + m}}}\)

Simplify the Ratio

We can now simplify this expression, canceling out common terms: \(\frac{f}{f'} = \frac{\sqrt{\frac{k}{M}}}{\sqrt{\frac{k}{M + m}}}\) Dividing two square roots is the same as taking the square root of the ratio: \(\frac{f}{f'} = \sqrt{\frac{\frac{k}{M}}{\frac{k}{M + m}}}\) Cancel out k in the numerator and denominator: \(\frac{f}{f'} = \sqrt{\frac{M + m}{M}}\) Which is equivalent to: \(\frac{f}{f'} = \sqrt{1 + \frac{m}{M}}\)

Choosing the Correct Option

Comparing this to the given options, we find that none of the options exactly match our final answer. It's important to double check our working and ensure we didn't miss any steps, but assuming we followed the correct process, it looks like there could be an error in the question or the given options.

Key concepts

Frequency Ratio in Simple Harmonic Motion

The concept of frequency ratio is crucial when analyzing changes in a mass-spring system undergoing simple harmonic motion (SHM). In such systems, the frequency of oscillation depends on the mass attached to the spring and the spring constant. Initially, our system has a mass, \(M\), and a spring constant, \(k\). The formula for the frequency \(f\) is given by:
\[f = \frac{1}{2 \pi} \sqrt{\frac{k}{M}}\]
When an additional mass, \(m\), is added to the system mid-motion, it alters the total mass to \(M + m\). Consequently, the new frequency \(f'\) after adding the mass becomes:
\[f' = \frac{1}{2 \pi} \sqrt{\frac{k}{M+m}}\]
By dividing the initial frequency by the new frequency, we derive the frequency ratio:
\[\frac{f}{f'} = \sqrt{\frac{M+m}{M}}\]
This expression reveals how the frequency decreases due to the added mass, reflecting an essential property in SHM where increased mass results typically in lower frequency of oscillation.

Mass-Spring System Details

A mass-spring system is a quintessential example of simple harmonic motion (SHM). Here, the spring exerts a restoring force that is proportional to the displacement from its equilibrium position. The system consists of:

• Mass (\(M\)): Which affects how quickly the system oscillates.
• Spring Constant (\(k\)): A measure of how stiff the spring is, influencing the restoring force.

The balance of these elements defines the dynamics of the motion, mathematical expressed by the frequency formula \(f = \frac{1}{2 \pi} \sqrt{\frac{k}{M}}\). This tells us that as long as the spring follows Hooke's law, the frequency is dependent only on the system's mass and spring stiffness.
When an external mass is added, the total mass changes, leading to a new frequency calculation with a similar formula but involving \(M + m\). This results in exciting shifts in the oscillating characteristics, exemplifying the sensitive nature of SHM systems to physical changes.

Amplitude Change in SHM

Amplitude in simple harmonic motion refers to the maximum displacement from the equilibrium position. While the initial problem deals with frequency changes due to mass addition, one should consider how amplitude can also change.
When an additional mass \(m\) is smoothly added at the equilibrium point, the initial energy is shared between \(M\) and \(m\), potentially altering the system's amplitude of SHM. If no external force is added during the mass change, energy conservation principles predict that shifts in mechanical energy distribution can cause amplitude modifications.

• Initial amplitude (\(A\)): Related to the beginning energy budget.
• Final amplitude (\(A'\)): Reflects energy distribution changes post-mass addition.

This shows the interconnected relationship between system dynamics and mechanical energy distribution, influencing amplitude as well as frequency in responses to parameter changes in the system.