
A pendulum-controlled clock is transferred from Earth to Moon. What would be the effect on the clock.
Answer
216k+ views
Hint:-We will use the time period relation of the simple pendulum with the gravity to find the effect on the clock. If the time period becomes less on the Moon than the Earth, then it means it will become slower otherwise it will become more faster.
Complete Step by Step Explanation:
The time period of a simple pendulum is given by,
$T = 2\pi \sqrt {\dfrac{L}{g}} $ , -----(1)
Where $T$ is the time period of one oscillation of pendulum,
$L$ is the length of pendulum
$g$ is the gravity in which pendulum is kept
Let $g$ be the gravity of Earth, while $g'$ be the gravity of moon
Then, time period on moon will be,
$T' = 2\pi \sqrt {\dfrac{L}{{g'}}} $ (since, the length of pendulum will be same everywhere)----(2)
Now, we know that gravity on Earth is six times the gravity on Moon, therefore,
$g = 6g'$
Using above relation in equation $1$
$T = 2\pi \sqrt {\dfrac{L}{{6g'}}} $
On rearranging above equation, we get,
$T = \dfrac{1}{{\sqrt 6 }}2\pi \sqrt {\dfrac{L}{{g'}}} $
Comparing above equation with equation $2$ , we get,
$T = \dfrac{1}{{\sqrt 6 }}T'$
So, we get,
$T' = \sqrt 6 T$
Now, we know, $\sqrt 6 > 2$
Therefore, the time period of a simple pendulum on the Moon will be more than two times that on the Earth. It means the pendulum will be oscillating slower on the Moon than on Earth.
Note:From the relation of the time period of a simple pendulum, it is clear that it’s time period doesn’t depend upon its mass and totally depends on gravity, that’s why we say, time runs slower on a planet with greater mass.
Complete Step by Step Explanation:
The time period of a simple pendulum is given by,
$T = 2\pi \sqrt {\dfrac{L}{g}} $ , -----(1)
Where $T$ is the time period of one oscillation of pendulum,
$L$ is the length of pendulum
$g$ is the gravity in which pendulum is kept
Let $g$ be the gravity of Earth, while $g'$ be the gravity of moon
Then, time period on moon will be,
$T' = 2\pi \sqrt {\dfrac{L}{{g'}}} $ (since, the length of pendulum will be same everywhere)----(2)
Now, we know that gravity on Earth is six times the gravity on Moon, therefore,
$g = 6g'$
Using above relation in equation $1$
$T = 2\pi \sqrt {\dfrac{L}{{6g'}}} $
On rearranging above equation, we get,
$T = \dfrac{1}{{\sqrt 6 }}2\pi \sqrt {\dfrac{L}{{g'}}} $
Comparing above equation with equation $2$ , we get,
$T = \dfrac{1}{{\sqrt 6 }}T'$
So, we get,
$T' = \sqrt 6 T$
Now, we know, $\sqrt 6 > 2$
Therefore, the time period of a simple pendulum on the Moon will be more than two times that on the Earth. It means the pendulum will be oscillating slower on the Moon than on Earth.
Note:From the relation of the time period of a simple pendulum, it is clear that it’s time period doesn’t depend upon its mass and totally depends on gravity, that’s why we say, time runs slower on a planet with greater mass.
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