Question

A mass M attached to a horizontal spring, executes SHM with amplitude ${{A}_{1}}$. When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude ${{A}_{2}}$. The ratio of $\left( \dfrac{{{A}_{1}}}{{{A}_{2}}} \right)$is:

& A.\,\dfrac{M}{M+m} \\\ & B.\,\dfrac{M+m}{M} \\\ & C.\,\sqrt{\dfrac{M}{M+m}} \\\ & D.\,\sqrt{\dfrac{M+m}{M}} \\\ \end{aligned}$$

Answer 1

We can solve this problem by either considering the time period of oscillations or the frequency of the oscillation. We will be using the formulae of time period of oscillation, the frequency and the conservation of momentum at the steady state to compute the ratio of the amplitudes.
Formula used:

& T=2\pi \sqrt{\dfrac{m}{k}} \\\ & \upsilon =A\omega \\\ & {{M}_{1}}{{v}_{1}}={{M}_{2}}{{v}_{2}} \\\ \end{aligned}$$ **Complete step by step answer:** From the given information, we have the data as follows. A mass M attached to a horizontal spring, executes SHM with amplitude $${{A}_{1}}$$. When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude $${{A}_{2}}$$. The formulae that we will be using are given as follows. The time period of oscillations. $$T=2\pi \sqrt{\dfrac{m}{k}}$$ Where m is the mass and k is the spring constant. The frequency of the oscillation. $$\upsilon =A\omega $$ Where A is the amplitude and w is the angular frequency. Consider the velocity of mass M be $${{v}_{1}}$$passing through the mean position and the velocity of mass $$(m+M)$$ be $${{v}_{2}}$$ passing through the mean position. Consider the time period of oscillation of these systems. $$\begin{aligned} & {{T}_{1}}=2\pi \sqrt{\dfrac{M}{k}} \\\ & {{T}_{2}}=2\pi \sqrt{\dfrac{(m+M)}{k}} \\\ \end{aligned}$$ Using the law of conservation of the linear momentum $$M{{v}_{1}}=(m+M){{v}_{2}}$$ Represent the above equation in terms of amplitude and the angular frequency. $$M\left( {{A}_{1}}{{\omega }_{1}} \right)=(m+M)\left( {{A}_{2}}{{\omega }_{2}} \right)$$ Represent the above equation in terms of the ratio of amplitudes. $$\dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{(m+M)}{M}\dfrac{{{\omega }_{2}}}{{{\omega }_{1}}}$$ Represent the above equation in terms of the time period. $$\begin{aligned} & \dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{(m+M)}{M}\dfrac{{{T}_{1}}}{{{T}_{2}}} \\\ & \Rightarrow \dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{(m+M)}{M}\dfrac{2\pi \sqrt{\dfrac{M}{k}}}{2\pi \sqrt{\dfrac{(m+M)}{k}}} \\\ & \therefore \dfrac{{{A}_{1}}}{{{A}_{2}}}=\sqrt{\dfrac{(m+M)}{M}} \\\ \end{aligned}$$ **$$\therefore $$ The ratio of the amplitudes is $$\dfrac{{{A}_{1}}}{{{A}_{2}}}=\sqrt{\dfrac{(m+M)}{M}}$$.** **Note:** At the mean position, the speed of the mass will be maximum and can be calculated using the formula that relates the frequency of oscillation, amplitude and the angular frequency. The spring constant remains the same in both the cases, so, gets cancelled out.