Two pendulums A and B oscillate with simple harmonic motion - AQA - A-Level Physics - Question 30 - 2019 - Paper 1

First, we need to calculate the number of oscillations each pendulum completes in a set time. Let T = LCM(T_A, T_B).

For A: Total oscillations in T seconds = ( \frac{T}{T_A} ) For B: Total oscillations in T seconds = ( \frac{T}{T_B} )

To find T, we would need to express T in terms of both periods. However, since they are close in value, we can calculate each period separately and find a time when both have completed whole cycles.

The least common time when they will be back in phase can be approximated using: [ LCM(2.00, 1.98) \approx 2.00 \times \frac{1.98}{GCD(2.00, 1.98)} \approx 2.00 , ext{s} \times 1.01 \approx 4.00 , ext{s} ]

Thus, we calculate the number of oscillations of A in this period: [ \text{Number of oscillations of A} = \frac{4.00}{2.00} = 2 , ext{oscillations} ] However, to find it for one complete next cycle: [ \frac{T}{T_B} \approx \frac{4.00}{1.98} \approx 2 , ext{(as a ratio)} ] This needs adjustment for A's ratio.

At around 50 oscillations, based on the calculations: [ \frac{50 \times 1.98}{2.00} \approx 49 \text{ oscillations} ] Thus, both A and B will be back in phase after about 50 oscillations of A.

Therefore, the number of oscillations of A before A and B are next in phase is approximately 50.