GENERAL
1. Moon's gravity is about 1/6 of that on earth, what is the period of a seconds pendulum on the Moon? ( Neglect thermal contraction/ expension)
A)2*√6 sec B) 2*√3 sec C) 2 sec
D) 3 sec
Solution:
A second pendulum on Earth has a period of 2 seconds
The period of simple pendulum :
T= 2*π *√(L/g)
-----------------------(1)
L- length of the pendulum (m)
g- acceleration due to gravity (gm/cc)
T- time period (sec)
The time period of simple pendulum (Earth) TE = 2*π *√(L/gE)
------------(2)
The time period of
simple pendulum ( Moon) Tm = 2*π *√(L/gm)-----------(3)
Dividing eq(2) by eq(3)
TE/ Tm = (
2*π*√
(L/gE) ) / ( 2*π √(L/gm)
)
TE/ Tm = (√ (1/gE) / √( 1/gm)
)
From above problem,
Suppose, Earth acceleration due to
gravity is =1
The moom’s acceleration due to gravity is = 1/6
TE/ Tm = ( √(
1/1 ) / √( 1/ (1/6) )
TE/ Tm = ( 1/√6)
Tm = √6 TE
Therefore, TE is 2
seconds
Tm= 2*√6
The period of the second pendulum
on the moon is =2*√6
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