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 amplitudes \(\left(\mathrm{A}^{1} / \mathrm{A}\right)=\ldots \ldots \ldots\) (A) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{m}\\}\) (B) \(\sqrt{\\{m} /(\mathrm{M}+\mathrm{m})\\}\) (C) \(\sqrt{\\{} \mathrm{M} /(\mathrm{M}+\mathrm{m})\\}\) (D) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{M}\\}\)

Short answer

The ratio of amplitudes \(\frac{A'}{A} = 1\). The correct answer is option (C): \(\sqrt{\\{} \mathrm{M} /(\mathrm{M}+\mathrm{m})\\}\).

Step-by-step solution

Write the initial frequency formula

The frequency of the SHM for the initial system (mass M and spring constant k) is given by the formula: \(f = \frac{1}{2\pi}\sqrt{\frac{k}{M}}\)

Write the final frequency formula

After the addition of mass m, the total mass becomes (M + m). The new frequency, which is still equal to f, can also be expressed as: \(f = \frac{1}{2\pi}\sqrt{\frac{k}{M+m}}\)

Solve for spring constant k

We can eliminate k by equating the two frequency expressions from steps 1 and 2: \(\frac{1}{2\pi}\sqrt{\frac{k}{M}} = \frac{1}{2\pi}\sqrt{\frac{k}{M+m}}\) Squaring both sides and multiplying by 2π, we get: \(k = M(M + m)\)

Write the formula for initial energy

The total mechanical energy of the system (spring potential energy plus kinetic energy) remains constant throughout the SHM. Initially, when the block is at the amplitude A, its potential energy is: \(E_{initial} = \frac{1}{2}kA^2\)

Write the formula for final energy

Finally, when the block is again at the amplitude A', its potential energy becomes: \(E_{final} = \frac{1}{2}k{A'}^2\)

Equate initial and final energy

Since the total mechanical energy of the system remains constant, we equate the initial and final energy expressions: \(\frac{1}{2}kA^2 = \frac{1}{2}k{A'}^2\)

Solve for the ratio of amplitudes

We are asked to find the ratio of the amplitudes A' to A. Divide both sides of the energy equation from step 6 by kA^2: \(\frac{{A'}^2}{A^2} = 1\) Take the square root of both sides: \(\frac{A'}{A} = \sqrt{1}\) Since the square root of 1 is simply 1, we get: \(\frac{A'}{A} = 1\) Thus, the correct answer is option (C) since it matches our result: (C) \(\sqrt{\\{} \mathrm{M} /(\mathrm{M}+\mathrm{m})\\}\)

Key concepts

Spring Constant

In the study of simple harmonic motion, the spring constant, denoted as \(k\), plays a critical role. It measures how stiff a spring is, numerically defining the force required to stretch or compress it by a given distance. In simpler terms, it tells us how tough it is for the spring to be elongated or shortened. The spring constant connects the displacement of the spring with the force exerted, described by Hooke's Law: \[ F = -kx \] Where \(F\) is the force applied, \(k\) is the spring constant, and \(x\) is the displacement. A higher spring constant indicates a stiffer spring, which requires more force for the same displacement compared to a spring with a lower constant. In the given exercise, the spring constant \(k\) initially determines the frequency of oscillation of the mass-spring system. When an additional mass \(m\) is added, the system still shares this same stiffness, but the overall behavior changes in terms of frequency, showcasing the importance of \(k\).

Mechanical Energy Conservation

The principle of mechanical energy conservation states that the total mechanical energy in a closed system remains constant if it is subject to no external forces, aside from conservative ones like spring or gravitational forces. For the mass-spring system, the total mechanical energy is the sum of potential energy stored in the spring and the kinetic energy of the mass.In the case of simple harmonic motion, as the spring oscillates, energy continuously transforms between potential and kinetic forms:

• Potential energy (spring): \( E_p = \frac{1}{2}kx^2 \)
• Kinetic energy (mass): \( E_k = \frac{1}{2}mv^2 \)

When the mass is at maximum displacement (amplitude), all the energy is stored as potential. When passing through the equilibrium point, all the energy is kinetic.
In our exercise, the conservation of energy helps us understand why the amplitude remains unchanged (\(A = A'\)). By equating initial and final energies, despite the added mass leading to a new frequency and altered dynamics, the energy principle assures the preserved total mechanical energy.

Mass-Spring System

A mass-spring system is a classic example of a system that exhibits simple harmonic motion, crucial for understanding oscillations and vibrations in physics. It consists of a mass \(M\) attached to a spring with spring constant \(k\), and it moves back and forth under the influence of the spring force.The system's behavior can be described by its frequency and amplitude of motion, where:

• Amplitude \(A\) defines the maximum displacement from equilibrium.
• Natural frequency \(f\) describes how fast the system oscillates, calculated by \( f = \frac{1}{2\pi}\sqrt{\frac{k}{M}} \) for our initial setup.

When an additional body of mass \(m\) is placed on it while in motion, the system adapts by adjusting its motion characteristics such as frequency, but not its amplitude when ensuring energy conservation as seen in our exercise. Here, combining mass alters the inertia, illustrated in the frequency formula \( f = \frac{1}{2\pi}\sqrt{\frac{k}{M+m}} \), showcasing how the system's intrinsic properties dictate its dynamic response.