A mass `m_(1)=1kg` connected to a horizontal spring performs S.H.M. with amplitude A. While mass `m_(1)` is passing through mean position another mass `m_(2)=3kg` is placed on it so that both the masses move together with amplitude `A_(1)`. The ratio of `(A_(1))/(A)` is `((p)/(q))^(1//2)`, where p and q are the smallest integers. Then what is the value of `p+q`?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Conditions
We have a mass \( m_1 = 1 \, \text{kg} \) connected to a spring, performing simple harmonic motion (SHM) with amplitude \( A \). The maximum velocity \( v \) of this mass when it is at the mean position is given by:
\[ v = \omega A \]
where \( \omega \) is the angular frequency.

### Step 2: Determine the Angular Frequency
The angular frequency \( \omega \) is related to the spring constant \( k \) and the mass \( m_1 \) as follows:
\[ \omega = \sqrt{\frac{k}{m_1}} = \sqrt{\frac{k}{1}} = \sqrt{k} \]

### Step 3: Introduce the Second Mass
When the mass \( m_2 = 3 \, \text{kg} \) is placed on \( m_1 \) at the mean position, the total mass becomes:
\[ m_{\text{total}} = m_1 + m_2 = 1 + 3 = 4 \, \text{kg} \]

### Step 4: Apply Conservation of Momentum
Using conservation of momentum, the initial momentum \( p_i \) must equal the final momentum \( p_f \):
\[ m_1 v = m_{\text{total}} v' \]
where \( v' \) is the new velocity after mass \( m_2 \) is added. Substituting the values, we have:
\[ 1 \cdot (\omega A) = 4 v' \]
From this, we can solve for \( v' \):
\[ v' = \frac{\omega A}{4} \]

### Step 5: Find the New Angular Frequency
The new angular frequency \( \omega' \) for the combined mass is:
\[ \omega' = \sqrt{\frac{k}{m_{\text{total}}}} = \sqrt{\frac{k}{4}} = \frac{1}{2} \sqrt{k} \]

### Step 6: Relate the New Amplitude to the Original Amplitude
The new amplitude \( A_1 \) can be expressed in terms of the new velocity:
\[ v' = \omega' A_1 \]
Substituting \( \omega' \):
\[ \frac{\omega A}{4} = \left(\frac{1}{2} \sqrt{k}\right) A_1 \]
This simplifies to:
\[ \frac{\omega A}{4} = \frac{1}{2} \sqrt{k} A_1 \]

### Step 7: Substitute \( \omega \) and Solve for \( A_1 \)
Substituting \( \omega = \sqrt{k} \):
\[ \frac{\sqrt{k} A}{4} = \frac{1}{2} \sqrt{k} A_1 \]
Dividing both sides by \( \sqrt{k} \):
\[ \frac{A}{4} = \frac{1}{2} A_1 \]
Multiplying both sides by 2:
\[ \frac{A}{2} = A_1 \]

### Step 8: Find the Ratio of Amplitudes
Now, we find the ratio of the new amplitude to the original amplitude:
\[ \frac{A_1}{A} = \frac{A/2}{A} = \frac{1}{2} \]

### Step 9: Express the Ratio in the Required Form
The problem states that:
\[ \frac{A_1}{A} = \left(\frac{p}{q}\right)^{1/2} \]
From our calculation, we have:
\[ \frac{1}{2} = \left(\frac{1}{4}\right)^{1/2} \]
Thus, we can identify \( p = 1 \) and \( q = 4 \).

### Step 10: Calculate \( p + q \)
Finally, we find:
\[ p + q = 1 + 4 = 5 \]

### Final Answer
The value of \( p + q \) is \( \boxed{5} \).

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