A mas m is moving in SHM on a line with amplitude A and frequency f. suddenly half of the mass comes to rest just at the moment when it comes mean position then the new amplitude becomes `lamda`A, then `lamda` will be

Options

A. 43832
B. `1/sqrt2`
C. `sqrt2`
D. 1

Answer

To solve the problem, we will analyze the situation step by step, using the principles of simple harmonic motion (SHM) and conservation of momentum.

### Step 1: Understanding the Initial Conditions
We have a mass \( m \) moving in simple harmonic motion (SHM) with an amplitude \( A \) and frequency \( f \). The mass oscillates due to a spring with spring constant \( k \).

**Hint:** Remember that the frequency of SHM is related to the mass and spring constant by the formula \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \).

### Step 2: Determine the Initial Angular Frequency
The angular frequency \( \omega_1 \) of the mass \( m \) is given by:
\[ \omega_1 = \sqrt{\frac{k}{m}} \]

**Hint:** The angular frequency is a measure of how quickly the mass oscillates.

### Step 3: Analyze the Change in Mass
When half of the mass \( (m/2) \) comes to rest at the mean position, the remaining mass that continues to oscillate is \( m/2 \). The new angular frequency \( \omega_2 \) for the remaining mass is:
\[ \omega_2 = \sqrt{\frac{k}{m/2}} = \sqrt{\frac{2k}{m}} = \sqrt{2} \cdot \sqrt{\frac{k}{m}} = \sqrt{2} \cdot \omega_1 \]

**Hint:** When the mass changes, the new angular frequency can be derived from the original frequency by considering the new effective mass.

### Step 4: Relate the Velocities at the Mean Position
At the mean position, the velocity of the mass can be expressed as:
\[ v_1 = \omega_1 A \]
for the original mass \( m \), and for the new mass \( m/2 \):
\[ v_2 = \omega_2 A_2 \]
where \( A_2 \) is the new amplitude we need to find.

**Hint:** The velocity at the mean position is maximum and is directly proportional to the amplitude.

### Step 5: Apply Conservation of Momentum
Since no external forces act on the system, we can apply the conservation of momentum. The momentum before half the mass comes to rest is equal to the momentum after:
\[ m v_1 = \frac{m}{2} v_2 \]
Substituting the velocities:
\[ m (\omega_1 A) = \frac{m}{2} (\omega_2 A_2) \]

**Hint:** The conservation of momentum principle states that the total momentum before an event must equal the total momentum after the event.

### Step 6: Substitute the Angular Frequencies
Substituting \( \omega_1 \) and \( \omega_2 \):
\[ m \left(\sqrt{\frac{k}{m}} A\right) = \frac{m}{2} \left(\sqrt{2} \cdot \sqrt{\frac{k}{m}} A_2\right) \]
This simplifies to:
\[ \sqrt{km} A = \frac{m}{2} \sqrt{2} \cdot \sqrt{\frac{k}{m}} A_2 \]

### Step 7: Cancel and Rearrange
Cancel \( m \) and \( \sqrt{k} \):
\[ A = \frac{1}{2} \sqrt{2} A_2 \]
Rearranging gives:
\[ A_2 = \frac{2A}{\sqrt{2}} = \sqrt{2} A \]

### Step 8: Find the Value of \( \lambda \)
From the problem, we know that:
\[ A_2 = \lambda A \]
Thus, we have:
\[ \lambda A = \sqrt{2} A \]
This leads to:
\[ \lambda = \sqrt{2} \]

### Final Answer
The value of \( \lambda \) is \( \sqrt{2} \).

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