Two particle A and B of equal masses are suspended from two massless springs of spring constants `k_(1)` and `k_(2)`, respectively. If the maximum velocities, during oscillations are equal, the ratio of amplitude of A and B is `(4//3) xx 1000 kg//m^(3)`. What relationship betwen `t` and `t_(0)` is ture?

Options

A. `sqrt(k_(1)//k_(2))`
B. `k_(1)//k_(2)`
C. `sqrt(k_(2)//k_(1))`
D. `k_(2)//k_(1)`

Explanation

To solve the problem, we need to analyze the relationship between the maximum velocities of two particles A and B, which are oscillating on springs with different spring constants.

### Step-by-Step Solution:

1. **Understanding Maximum Velocity in SHM**:
The maximum velocity \( V_{max} \) of a particle in simple harmonic motion (SHM) is given by the formula:
\[ V_{max} = \omega A \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude.

2. **Angular Frequency**:
The angular frequency \( \omega \) for a mass-spring system is defined as:
\[ \omega = \sqrt{\frac{k}{m}} \]
where \( k \) is the spring constant and \( m \) is the mass of the particle.

3. **Setting Up the Equations**:
For particles A and B, we can write:
\[ V_{max, A} = \omega_A A_A = \sqrt{\frac{k_1}{m}} A_A \]
\[ V_{max, B} = \omega_B A_B = \sqrt{\frac{k_2}{m}} A_B \]
Given that the maximum velocities are equal, we can set these two equations equal to each other:
\[ \sqrt{\frac{k_1}{m}} A_A = \sqrt{\frac{k_2}{m}} A_B \]

4. **Cancelling Mass**:
Since the masses of the particles are equal, we can cancel \( m \) from both sides:
\[ \sqrt{k_1} A_A = \sqrt{k_2} A_B \]

5. **Finding the Ratio of Amplitudes**:
Rearranging gives us:
\[ \frac{A_A}{A_B} = \frac{\sqrt{k_2}}{\sqrt{k_1}} \]

6. **Relating Amplitudes to Given Ratio**:
We are given that the ratio of amplitudes \( \frac{A_A}{A_B} = \frac{4}{3} \times 1000 \, \text{kg/m}^3 \). This implies:
\[ \frac{A_A}{A_B} = \frac{4}{3} \times 1000 \]

7. **Finding the Relationship between Periods**:
The period \( T \) of a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Therefore, the periods for A and B are:
\[ T_A = 2\pi \sqrt{\frac{m}{k_1}}, \quad T_B = 2\pi \sqrt{\frac{m}{k_2}} \]

8. **Finding the Ratio of Periods**:
The ratio of the periods is:
\[ \frac{T_A}{T_B} = \sqrt{\frac{k_2}{k_1}} \]

9. **Final Relationship**:
Since we have \( \frac{A_A}{A_B} = \frac{\sqrt{k_2}}{\sqrt{k_1}} \), we can conclude that:
\[ T_A \propto \sqrt{\frac{m}{k_1}}, \quad T_B \propto \sqrt{\frac{m}{k_2}} \]
Thus, the relationship between \( T \) and \( T_0 \) (where \( T_0 \) is a reference period) is:
\[ T_A = T_0 \sqrt{\frac{k_2}{k_1}} \]

### Conclusion:
The relationship between \( T \) and \( T_0 \) is determined by the ratio of the spring constants \( k_1 \) and \( k_2 \).