Two particles `P_(1) and P_(2)` having identical masses are suspended from two massless springs of spring constant `k_(1) ` = 5 N/m and `k_(2)` = 8 N/m respectively. If maximum velocity of oscillation for particle `P_(1)` is twice than for particle `P_(2)`, then find the ratio of amplitudes for `P_(1)` and `P_(2)`.

Explanation

To solve the problem, we need to find the ratio of the amplitudes of two particles \( P_1 \) and \( P_2 \) suspended from springs with different spring constants. We know that the maximum velocity of oscillation for \( P_1 \) is twice that of \( P_2 \).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Spring constant for \( P_1 \): \( k_1 = 5 \, \text{N/m} \)
- Spring constant for \( P_2 \): \( k_2 = 8 \, \text{N/m} \)
- The maximum velocity of \( P_1 \) is twice that of \( P_2 \): \( V_{1_{\text{max}}} = 2 V_{2_{\text{max}}} \)
...