Question

two simple pendulums of length m and m are at their mean position with their velocities in the same direction at some instant. after how many oscillations of bigger pendulum they will again be in the same phase?

Answer 1

The big pendulum swings x times, and then it's in phase with the small pendulum.

How to solve this:

Let T1 be the time it takes for the smaller pendulum (with length m) to swing back and forth.

Let T2 represent the time it takes for the longer pendulum (measured in meters) to complete one swing.

The ratio of the time period:

We can compare the time it takes for a pendulum to swing back and forth using a formula: T = 2π * √(l/g), where l is the length of the pendulum and g is the force of gravity (about 9. 81 m/s²)

Hence, T2/T1 = √(m/m) = 1 (since both lengths are equal).

Replace the time period ratio from step 2 with xT1 = y(1T1) = xT1.

This equation becomes x = y.

So, the shorter pendulum also has to do x swings for both pendulums to be in phase again.