Question 225: Oscillations

Question

A mass ' $M$ ' attached to a horizontal spring executes S.H.M. of 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 $\mathrm{A}_2$. The ratio $\left(\frac{A_1}{A_2}\right)$ is

Accepted Answer

Step 1: Conservation of Momentum at Mean Position
At the mean position, the net force ($\mathrm{F}_{	ext{net}}$) is zero. This means the total momentum before and after adding mass must stay the same.
Step 2: Setting Up the Momentum Equation
Before the smaller mass $m$ is placed, only mass $M$ moves, so its momentum is $M v_1$. After adding $m$, both move together, so their momentum is $(M+m) v_2$.
$$M v_1 = (M+m) v_2$$
Step 3: Express Velocity in Terms of Amplitude and Angular Frequency
The velocity at mean position for a mass in S.H.M. is $V = A\omega$. For $A_1$ and mass $M$, $v_1 = A_1 \omega_1$. For $A_2$ and mass $(M+m)$, $v_2 = A_2 \omega_2$.
$$M \omega_1 A_1 = (M+m) \omega_2 A_2$$
Step 4: Find Angular Frequencies for Both Cases
The angular frequency $\omega = \sqrt{\frac{k}{m}}$, where $k$ is the spring constant and $m$ is the mass.
$$\omega_1 = \sqrt{\frac{k}{M}} \quad 	ext{and} \quad \omega_2 = \sqrt{\frac{k}{M+m}}$$
Step 5: Substitute and Solve for the Ratio $ \frac{A_1}{A_2} $
Replace $\omega_1$ and $\omega_2$ in the momentum equation:
$$M \sqrt{\frac{k}{M}} A_1 = (M+m) \sqrt{\frac{k}{M+m}} A_2$$
Step 6: Simplify the Equation
Simplifying both sides, we get:
$$\frac{A_1}{A_2} = \sqrt{\frac{M+m}{M}}$$

Answer

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

Answer

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

Answer

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

Answer

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

Oscillations

A mass ' M ' attached to a horizontal spring executes S.H.M. of 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 \mathrm{A}_2. The ratio \left(\frac{A_1}{A_2}\right) is

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