Both will ring together after time 4T. (integral LCM of 1 and 5/4 is 4)

Introduction:
In this scenario, we have two simple pendulums with different time periods, denoted by T and 5T/4. We are given that these pendulums are in phase at a certain time and we are required to determine the time it takes for them to be in phase again.

Analysis:
To solve this problem, we need to understand the concept of phase difference between two oscillating systems. The phase difference is the fraction of a cycle by which one oscillation lags behind or leads another oscillation.

Given:
Time period of pendulum 1, T1 = T
Time period of pendulum 2, T2 = 5T/4

Phase difference:
Phase difference (Δφ) can be calculated using the formula:
Δφ = (T2 - T1) / T1

Let's substitute the given values:
Δφ = (5T/4 - T) / T

Simplifying the expression:
Δφ = (5/4 - 1) = 1/4

This means that the phase difference between the two pendulums is 1/4 cycle.

Finding the time to be in phase again:
To determine the time it takes for the two pendulums to be in phase again, we need to find the time required for the phase difference to be an integer multiple of 2π radians or a whole number of cycles.

Let's assume the time required for the pendulums to be in phase again is T' (in seconds).

The phase difference after time T' can be calculated as:
Δφ' = (T2 - T1) / T'

Since we want the pendulums to be in phase again, the phase difference Δφ' should be equal to an integer multiple of 2π radians.

Mathematically, we can express this as:
Δφ' = 2π * n, where n is an integer

Substituting the given values:
(5T/4 - T) / T' = 2π * n

Simplifying the expression:
5T/4 - T = 2π * n * T'

Rearranging the equation to solve for T':
T' = (5T/4 - T) / (2π * n)

Answer:
From the above equation, we can see that the time required for the two pendulums to be in phase again is given by T' = (5T/4 - T) / (2π * n), where n is an integer.
In this case, the possible values for T' are:
1) When n = 4, T' = (5T/4 - T) / (2π * 4) = 4T
2) When n = 3, T' = (5T/4 - T) / (2π * 3)
3) When n = 6, T' = (5T/4 - T) / (2π * 6)
4) When n = 5, T' = (5T/4 - T) / (2π * 5)

Therefore, the possible times for the two pendulums to be in phase again are 4T, 3T