Question

The equation that governs the period of a pendulum’s swinging. T=2π√L/g

Where T is the period, L is the length of the pendulum and g is a constant, equal to 9.8 m/s2. The symbol g is a measure of the strength of Earth’s gravity, and has a different value on other planets and moons.


On our Moon, the strength of earth’s gravity is only 1/6th of the normal value. If a pendulum on Earth has a period of 4.9 seconds, what is the period of that same pendulum on the moon?

Answer 1

The period of that same pendulum on the moon is 12.0 s

Step-by-step explanation:

Given;

period of a pendulum’s swinging, T=2π√L/g

the strength of earth’s gravity on moon, g₂ = ¹/₆(g₁)

period of pendulum on Earth, T₁ = 4.9 s

period of pendulum on moon, T₂ = ?

The length of the pendulum is constant, make it the subject of the formula;

`T = 2\pi \sqrt{(L)/(g) }\\\\(T)/(2\pi) = \sqrt{(L)/(g)}\\\\((T)/(2\pi) )^2 =(L)/(g)\\\\(T^2)/(4\pi^2) = (L)/(g)\\\\ L = (gT^2)/(4\pi^2)\\\\L_1 = L_2\\\\(g_1T_1^2)/(4\pi^2)= (g_2T_2^2)/(4\pi^2)\\\\g_1T_1^2 = g_2T_2^2\\\\T_2^2 = (g_1T_1^2)/(g_2) \\\\T_2 = \sqrt{(g_1T_1^2)/(g_2)}\\\\ T_2 = \sqrt{(g_1T_1^2)/(g_1/6)}\\\\ T_2 = \sqrt{(6*g_1T_1^2)/(g_1)}\\\\T_2 = √(6T_1^2)\\\\ T_2 = T_1√(6) \\\\T_2 = (4.9)√(6)\\\\ T_2 = 12.0 \ s`

Therefore, the period of that same pendulum on the moon is 12.0 s

Answer 2

The period of that same pendulum on the moon is 12.0 seconds.

Step-by-step explanation:

To determine the period of that same pendulum on the moon,

First, we will determine the value of g (which is a measure of the strength of Earth's gravity) on the Moon. Let the value of g on the Moon be
`g_(M)`.

From the question, the strength of earth’s gravity is only 1/6th of the normal value. The normal value of g is 9.8 m/s²

∴
`g_(M)` =
`(1)/(6) * 9.8 m/s^(2)`

`g_(M)` = 1.63 m/s²

From the question, T=2π√L/g

`T = 2\pi \sqrt{(L)/(g) }`

We can write that,

`T_(E) = 2\pi \sqrt{(L)/(g_(E) ) }` .......... (1)

Where
`T_(E)` is the period of the pendulum on Earth and
`g_(E)` is the measure of the strength of Earth's gravity

and

`T_(M) = 2\pi \sqrt{(L)/(g_(M) ) }` .......... (2)

Where
`T_(M)` is the period of the pendulum on Moon and
`g_(M)` is the measure of the strength of Earth's gravity on the Moon.

Since we are to determine the period of the same pendulum on the moon, then,
`2\pi` and
`L` are constants.

Dividing equation (1) by (2), we get

`(T_(E) )/(T_(M) ) = \sqrt{(g_(M) )/(g_(E) ) }`

From the question,

`T_(E) = 4.9secs`

`g_(E)` = 9.8 m/s²

`g_(M)` = 1.63 m/s²

`T_(M)` = ??

From,

`(T_(E) )/(T_(M) ) = \sqrt{(g_(M) )/(g_(E) ) }`

`(4.9)/(T_(M) ) = \sqrt{(1.63)/(9.8) }`

`(4.9)/(T_(M) ) = 0.40783`

`T_(M) =(4.9)/(0.40783 )`

`T_(M) = 12.01 secs`

∴
`T_(M) = 12.0secs`

Hence, the period of that same pendulum on the moon is 12.0 seconds.