A mass M, attached to a horizontal spring, executes S.H.M. with amplitude A1. 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 A2. The ratio of is :

Engineering

Physics

Basics of Simple Harmonic Motion

Question

A mass M, attached to a horizontal spring, executes S.H.M. with amplitude A₁. 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 A₂. The ratio of $(\frac{A_{1}}{A_{2}})$ is :

${(\frac{M}{M + m})}^{1 / 2}$

${(\frac{M + m}{M})}^{1 / 2}$

$\frac{M + m}{M}$

$\frac{M}{M + m}$

C.O.L.M. MV_max = (m + M)V_new , V_max = A₁w₁

_{${V_{new}} = \frac{{M{V_{\max }}}}{{(m + M)}}$}

Now, V_new = A₂.w₂

_{$\frac{{M.{A_1}}}{{(m + M)}}\sqrt {\frac{K}{M}} = {A_2}\sqrt {\frac{K}{{(m + M)}}} $}

_{${A_2} = {A_1}\sqrt {\frac{M}{{(m + M)}}} $ $\frac{{{A_1}}}{{{A_2}}} = {\left( {\frac{{m + M}}{M}} \right)^{1/2}}$}

Concept Explanation: Simple Harmonic Motion with Mass Addition

When a mass executes Simple Harmonic Motion (SHM), its total energy remains conserved if no external forces act. At the mean position, the energy is purely kinetic. When a smaller mass is added at this point, momentum conservation applies since no external horizontal forces act.

Step-by-Step Solution:

Step 1: Determine the velocity at mean position before adding mass.

For SHM, maximum velocity $v_{max} = A ω$ . With amplitude A₁ and angular frequency $ω_{1} = \sqrt{\frac{k}{M}}$ , velocity $v_{1} = A_{1} \sqrt{\frac{k}{M}}$ .

Step 2: Apply conservation of momentum when mass m is added.

Initial momentum: $M \times v_{1}$ . Final momentum: $(M + m) v_{2}$ . So, $M v_{1} = (M + m) v_{2}$ .

Step 3: Relate the new maximum velocity to the new amplitude.

After adding mass, angular frequency becomes $ω_{2} = \sqrt{\frac{k}{M + m}}$ . Maximum velocity $v_{2} = A_{2} ω_{2} = A_{2} \sqrt{\frac{k}{M + m}}$ .

Step 4: Substitute and solve for the amplitude ratio.

From momentum conservation: $M A_{1} \sqrt{\frac{k}{M}} = (M + m) A_{2} \sqrt{\frac{k}{M + m}}$ .

Simplify: $M A_{1} \frac{1}{\sqrt{M}} = (M + m) A_{2} \frac{1}{\sqrt{M + m}}$ .

Which gives: $\frac{A_{1}}{A_{2}} = \frac{M}{+} \times \frac{\sqrt{M}}{\sqrt{M + m}} = \sqrt{\frac{M}{+}}$ .

Final Answer: The ratio $\frac{A_{1}}{A_{2}} = {(\frac{M + m}{M})}^{1 / 2}$ .

Key Formulae:

Angular Frequency: $ω = \sqrt{\frac{k}{m}}$
Maximum Velocity in SHM: $v_{max} = A ω$
Momentum Conservation: $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

Detailed Solution

Concept Explanation: Simple Harmonic Motion with Mass Addition

Step-by-Step Solution:

Related Topics:

Key Formulae:

Detailed Solution

Concept Explanation: Simple Harmonic Motion with Mass Addition

Step-by-Step Solution:

Related Topics:

Key Formulae:

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