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A pendulum with a ball of mass $m$ hanging from a string of length lis set in motion on Earth, and the system is found to have a frequency of $f$ . If the length of the string were doubled, the hanging mass tripled, and the system moved to the moon, what would be the new frequency?NOTE: Acceleration due to gravity of the Moon is approximately $\frac{1}{6}$ ofEarth's.(A) $\frac{1}{12} f$(B) $\sqrt{\frac{1}{12}} f$(C) $\sqrt{12} f$(D) 12$f$

B

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Chapter 1

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Section 1

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here we have frequency of a pendulum on Earth and then the length of the string is doubled. The mass is triple and the system is moved to the moon. Where extradition due to gravity is one x 6th of the Earth's gravity. And we have to find out new frequency on the moon. So on art the frequency of the pendulum is given by difficult to one x 2 by Mhm. G. Bye. And where G. Is X. Listen due to gravity on earth and Ellis lent now on moon it is said the frequencies F. M. And acceleration due to gravity is GM. And lent is doubled. So the new land will be too old. Now exhibition to to gravity GM is Equal to inflation due to gravity on earth by six. So we will substitute the value by two pi G by six two L. This gives us one by to buy G by to help help. So this means one by who attended 12 into one by two by Lieutenant G by L. And this is frequency on this portion is frequency on art. So we conducted so frequency on Moon is equal to one by 12 good, if hence, from the given options. The correct option is B. This completes the solution. Thank you.

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