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) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{M}\\}\)

Step by step solution

01

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: \(f = \frac{1}{2\pi}\sqrt{\frac{k}{M}}\)

02

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: \(f = \frac{1}{2\pi}\sqrt{\frac{k}{M+m}}\)

03

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: \(k = M(M + m)\)

04

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: \(E_{initial} = \frac{1}{2}kA^2\)

05

Write the formula for final energy

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

06

Equate initial and final energy

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

07

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) \(\sqrt{\\{} \mathrm{M} /(\mathrm{M}+\mathrm{m})\\}\)

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