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

Answer:

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

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

∴ $g_{M}$ = $\frac{1}{6} \times 9.8 m/s^{2}$

$g_{M}$ = 1.63 m/s²

From the question, T=2π√L/g

$T = 2\pi \sqrt{\frac{L}{g} }$

We can write that,

$T_{E} = 2\pi \sqrt{\frac{L}{g_{E} } }$ .......... (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

$T_{M} = 2\pi \sqrt{\frac{L}{g_{M} } }$ .......... (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

$\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }$

From the question,

$T_{E} = 4.9secs$

$g_{E}$ = 9.8 m/s²

$g_{M}$ = 1.63 m/s²

$T_{M}$ = ??

From,

$\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }$

$\frac{4.9}{T_{M} } = \sqrt{\frac{1.63}{9.8} }$

$\frac{4.9}{T_{M} } = 0.40783$

$T_{M} =\frac{4.9}{0.40783 }$

$T_{M} = 12.01 secs$

∴ $T_{M} = 12.0secs$

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

Answer 2

Answer:

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

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;

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

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