Question

Hlo users !

Q, (1) Find the length of second pendulum

(a) On moon

(b) on moon .
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Q.2 , A simple pendulum has time period t .find the length of another pendulum whose time period is 20% more than t. length of 1st pendulum is L.

Answer 1

$\bf{hello \: users}$

Solution

:-
$t \: = 2\pi \sqrt{ \frac{l}{g} }$

$2 = 2\pi \sqrt{ \frac{l}{g} }$

$1 = \pi \sqrt{ \frac{l}{g} }$

$l = \frac{g}{\pi {}^{2} }$

Solution(a) :-

on earth .

$on \: earth \: g \: = 9.8$

$l \: = \frac{9.8}{\pi {}^{2} }$
similar to 1 m .
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Solution (b)

on moon :-

g = 9.8/6

$l \: = \frac{9.8}{6\pi {}^{2} }$
Similar to 1/6m

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Q.2 :-

$t \: = 2\pi \sqrt{ \frac{l}{g} }$
$t1 = 2\pi \sqrt{ \frac{l1}{g} }$

$\frac{t1}{t} = \sqrt{ \frac{l1}{l} }$

$1.2 = \sqrt{ \frac{l1}{l} }$

$l1 = 1.44l$
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Hope it helps !!