Question

Two oscillating simple pendulums with time periods \(T\) and \(\frac{4T}{3}\) are in phase at a given time. They will be again in phase after an elapse of time

• A. \(5T\)
• B. \(4T\)
• C. \(3T\)
• D. \(2T\)

Hint:

They are in phase again when the shorter period has completed one more oscillation than the longer.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Two oscillators are in phase when each has completed a whole number of oscillations at the same instant. The first time this happens after the start is the least common multiple of their periods.

Step 2: Set up:
Periods are \(T\) and \(\frac{4T}{3}\). We need \(nT = m\cdot\frac{4T}{3}\) for the smallest whole numbers \(n, m\).

Step 3: Solve:
\(\frac nm = \frac43\), so \(n = 4\), \(m = 3\). The time elapsed is \(4T\).

Step 4: Check:
In time \(4T\) the first pendulum completes 4 oscillations. The second has period \(\frac{4T}{3}\), so it completes \(\frac{4T}{4T/3} = 3\) oscillations. Both are whole numbers.

Step 5: Why the other options are wrong.
At \(2T\) and \(3T\), the second pendulum has completed \(1.5\) and \(2.25\) oscillations. At \(5T\) it has completed \(3.75\).

Final Answer:
They are in phase again after \(4T\), option (B). \[ \boxed{4T} \]