Physics, asked by javeriailyas2022, 5 months ago

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 sereenasunil
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 ahkofficial10
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

Similar questions