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 step by step, we will analyze the situation involving simple harmonic motion (SHM) and the effects of halving the mass at the mean position.

### Step 1: Understand the initial conditions
The mass \( m \) is moving in SHM with amplitude \( A \) and frequency \( f \). The angular frequency \( \omega_1 \) can be expressed as:
\[ \omega_1 = 2\pi f = \sqrt{\frac{k}{m}} \]
where \( k \) is the spring constant.

### Step 2: Analyze the situation when half the mass comes to rest
When half of the mass comes to rest at the mean position, the effective mass that continues to oscillate is \( \frac{m}{2} \). The new angular frequency \( \omega_2 \) for the remaining mass can be calculated as:
\[ \omega_2 = \sqrt{\frac{k}{\frac{m}{2}}} = \sqrt{\frac{2k}{m}} = \sqrt{2} \cdot \omega_1 \]

### Step 3: Apply conservation of momentum
At the mean position, we apply the conservation of momentum. The momentum before half the mass comes to rest is equal to the momentum after:
\[ m v_1 = \frac{m}{2} v_2 \]
where \( v_1 \) is the velocity of the mass \( m \) at the mean position, and \( v_2 \) is the velocity of the mass \( \frac{m}{2} \) after half of it comes to rest.

### Step 4: Calculate velocities
The velocity \( v_1 \) at the mean position can be expressed in terms of amplitude and angular frequency:
\[ v_1 = A \omega_1 \]
For the remaining mass \( \frac{m}{2} \), the new amplitude \( A_2 \) can be expressed as:
\[ v_2 = A_2 \omega_2 \]

### Step 5: Substitute and simplify
Substituting the expressions for \( v_1 \) and \( v_2 \) into the momentum equation gives:
\[ m (A \omega_1) = \frac{m}{2} (A_2 \omega_2) \]
Cancelling \( m \) from both sides and substituting \( \omega_2 = \sqrt{2} \cdot \omega_1 \):
\[ A \omega_1 = \frac{1}{2} A_2 (\sqrt{2} \cdot \omega_1) \]
Cancelling \( \omega_1 \) from both sides (assuming \( \omega_1 \neq 0 \)):
\[ A = \frac{1}{2} A_2 \sqrt{2} \]

### Step 6: Solve for the new amplitude \( A_2 \)
Rearranging the equation gives:
\[ A_2 = \frac{2A}{\sqrt{2}} = \sqrt{2} A \]

### Step 7: Determine \( \lambda \)
Since it is given that the new amplitude is \( \lambda A \), we can equate:
\[ \lambda A = \sqrt{2} A \]
Thus, we find:
\[ \lambda = \sqrt{2} \]

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

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