A simple pendulum completes 6 vibrations in 12 seconds on the surface of earth, find its timeperiod on suface of Jupiter where acceleration due to gravity is 2.63 times that of Earth.
Answers
Answered by
1
Answer:
Acceleration due to gravity on the surface of the moon is
6
1
that of surface of the earth.
∴g
m
=
6
1
g
e
On earth, T
e
=2π
g
e
L
On Moon, T
m
=2π
g
m
L
∴
T
e
T
m
=
g
m
g
e
=
6
T
m
=
6
T
e
=
6
×4s=10s
Answered by
0
Answer:
Time Period of simple pendulum on the surface of Jupiter is 1.233 sec.
Explanation:
Time period on the surface of earth is:
T = time taken/no. of vibrations
T = 12/6 = 2 sec
Gravity of Jupiter = 2.63ge
We know that,
T = 2π√L/ge --- eq(i)
T' = 2π√L/gj
Since, gj = 2.63ge
T' = 2π√L/2.63ge
T' = 2π√L/ge × √1/2.63
T' = (2π√L/ge) × √1/2.63
From eq(i)
T' = T × √1/2.63
T' = 2 × √1/2.63
T' = 1.233 sec
Where,
T = Time Period on Earth
T' = Time Period on Jupiter
ge = Gravity of Earth
gj = Gravity of Jupiter
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