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: Phase difference approach:
In time $t$ the phases are $\frac{2\pi t}{T}$ and $\frac{2\pi t\cdot3}{4T}$. Their difference must be $2\pi$.

Step 2: Equation:
$2\pi t\left(\frac1T - \frac{3}{4T}\right) = 2\pi$, so $\frac{t}{4T} = 1$.

Step 3: Result:
$t = 4T$.

Final Answer:
The time is 4T, option (B). \[ \boxed{4T} \]