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```        At moon the weight of things become 1/6th of weight of earth . What is the ratio of time period of simple
pendulum at earth to that on the moon.```
6 years ago

510 Points
```										W = mg
Wmoon = Wearth/6
gmoon = gearth/6
T(time period) = 2pi(L/g)1/2
Tearth/Tmoon  = (gmoon/gearth)1/2
g(moon) = g(earth/6) so
Tearth/Tmoon = (1/6)1/2
```
6 years ago
Sudheesh Singanamalla
114 Points
```										Consider a simple pendulum of mass 'm' and length 'L' .
time period of a simple pendulum = 2*pi* sqrt( L/g ) ----------------------------- (on the earth) -------- (i)
time period of the simple pendulum = 2*pi* sqrt (6L / g ) ------------------------ (on the moon) -------- (ii)

ratio of time period of simple pendulum of earth to that on moon = (i) / (ii)
= 2*pi* sqrt( L/g) / 2*pi* sqrt ( 6L/g)
= sqrt (L/g ) / sqrt ( 6L/g )
= (L/g)/(6L/g)
= 1/6

therefore the ratio of timeperiod of the pendulum on the earth to that of the moon is 1:6 .
```
6 years ago
Ritvik Gautam
85 Points
```										We know, time period of a pendulum,
T=2π√l/g

Hence, substituting the values and using formula,
we get,
Tmoon = √6 x Tearth

```
6 years ago
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