Two pendulums of length $121 ext{ cm}$ and $100 ext{ cm}$ start vibrating in phase. At some instant, the t

Question

Two pendulums of length $121 	ext{ cm}$ and $100 	ext{ cm}$ start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

Accepted Answer

1.  **Time Period Relationship:** The time period ($T$) of a simple pendulum is directly proportional to the square root of its length ($L$): $T \propto \sqrt{L}$.
2.  **Given Data:**
    *   Length of longer pendulum ($L_L$) = $121 	ext{ cm}$
    *   Length of shorter pendulum ($L_S$) = $100 	ext{ cm}$
3.  **Ratio of Time Periods:**
    $$ \frac{T_L}{T_S} = \sqrt{\frac{L_L}{L_S}} = \sqrt{\frac{121}{100}} = \frac{11}{10} $$
    This implies $10 T_L = 11 T_S$. That is, the time taken for 10 oscillations of the longer pendulum is equal to the time taken for 11 oscillations of the shorter pendulum.
4.  **Phase Condition:** Two oscillating bodies starting in phase will be in phase again when the faster body (shorter pendulum) completes one more full oscillation than the slower body (longer pendulum). Let $n$ be the number of vibrations of the shorter pendulum.
    $$ n T_S = (n-k) T_L $$
    For the first time they come in phase, the difference in number of oscillations is usually related to the ratio integers. From the ratio $T_L/T_S = 11/10$, the smallest integers satisfying the condition $n_L T_L = n_S T_S$ are $n_L = 10$ and $n_S = 11$.
5.  **Conclusion:** The shorter pendulum completes 11 vibrations to be in phase again.

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