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Two apdulums begin to swing simultaneosu...

Two apdulums begin to swing simultaneosuly. The first pendulum makes `9` full oscillations when the other makes `7`. Find the ratio of length of the two pendulums

A

`(49)/(81)`

B

`(7)/(9)`

C

`(50)/(81)`

D

`(1)/(2)`

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To solve the problem of finding the ratio of the lengths of two pendulums based on their oscillations, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Problem**: - We have two pendulums. The first pendulum completes 9 oscillations while the second pendulum completes 7 oscillations in the same time period. 2. **Defining Variables**: - Let \( T_1 \) be the time period of the first pendulum. - Let \( T_2 \) be the time period of the second pendulum. - Let \( L_1 \) be the length of the first pendulum. - Let \( L_2 \) be the length of the second pendulum. 3. **Relating Time and Oscillations**: - The total time taken by the first pendulum for 9 oscillations is \( 9T_1 \). - The total time taken by the second pendulum for 7 oscillations is \( 7T_2 \). - Since both pendulums start swinging simultaneously and complete their respective oscillations in the same total time, we can equate the two: \[ 9T_1 = 7T_2 \] 4. **Finding the Ratio of Time Periods**: - Rearranging the equation gives: \[ \frac{T_1}{T_2} = \frac{7}{9} \] 5. **Using the Relationship Between Time Period and Length**: - The time period of a pendulum is related to its length by the formula: \[ T \propto \sqrt{L} \] - Therefore, we can write: \[ T_1 \propto \sqrt{L_1} \quad \text{and} \quad T_2 \propto \sqrt{L_2} \] 6. **Setting Up the Ratio of Lengths**: - From the proportionality, we can express the lengths in terms of the time periods: \[ \frac{\sqrt{L_1}}{\sqrt{L_2}} = \frac{T_1}{T_2} \] - Substituting the ratio we found: \[ \frac{\sqrt{L_1}}{\sqrt{L_2}} = \frac{7}{9} \] 7. **Squaring Both Sides**: - To find the ratio of lengths, we square both sides: \[ \frac{L_1}{L_2} = \left(\frac{7}{9}\right)^2 = \frac{49}{81} \] 8. **Conclusion**: - Thus, the ratio of the lengths of the two pendulums is: \[ \frac{L_1}{L_2} = \frac{49}{81} \] ### Final Answer: The ratio of the lengths of the two pendulums is \( \frac{49}{81} \). ---
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RESONANCE-SIMPLE HARMONIC MOTION -Exercise- 1, PART - II
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  9. A particle moves on y-axis according to the equation y = A +B sin omeg...

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  10. Two particles execute SHMs of the same amplitude and frequency along t...

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  11. A mass M is performing linear simple harmonic motion. Then correct gra...

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  12. A body executing SHM passes through its equilibrium. At this instant, ...

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  13. The KE and PE , at is a particle executing SHM with amplitude A will b...

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  14. A point particle if mass 0.1 kg is executing SHM of amplitude 0.1 m. W...

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  15. For a particle performing SHM

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  16. Acceleration a versus time t graph of a body in SHM is given by a curv...

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