Question

Two pendulums of time periods \(3\,s\) and \(7\,s\), respectively, start oscillating simultaneously from opposite extreme positions. After how much time will they be in same phase?

• A. \( \frac{21}{8}\,s \)
• B. \( \frac{21}{4}\,s \)
• C. \( \frac{21}{2}\,s \)
• D. \( \frac{21}{10}\,s \)

Hint:

Opposite extreme start $\Rightarrow$ initial phase difference \(=\pi\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Two particles starting from opposite extremes have an initial phase difference of \(\pi\). They come into the same phase when their relative phase difference changes by an additional \(\pi\) (making the total difference \(2\pi\)).
Step 2: Key Formula or Approach:
Phase \(\phi = \omega t + \phi_{0}\).
Relative phase \(\Delta\phi = (\omega_{1} - \omega_{2})t + \text{Initial phase difference}\).
: Detailed Explanation:
Let \(\omega_{1} = \frac{2\pi}{3}\) and \(\omega_{2} = \frac{2\pi}{7}\).
Initial phase difference \(= \pi\) (opposite extremes).
For the same phase:
\[ (\omega_{1} - \omega_{2})t = \pi \]
\[ \left( \frac{2\pi}{3} - \frac{2\pi}{7} \right)t = \pi \]
\[ 2\pi \left( \frac{7 - 3}{21} \right)t = \pi \]
\[ 2 \times \frac{4}{21} t = 1 \]
\[ \frac{8}{21} t = 1 \implies t = \frac{21}{8}\text{ s} \]
Step 3: Final Answer:
The required time is \(21/8\text{ s}\).