A second's pendulum is taken on the surface of moon where acceleration due to gravity is 1/6th of that of earth.Will the time period of pendulum remain same or increase or decrease?Give a reason.
take a look at the formula which gives the pendulum's period. It involves g, right?
Look what happens if you replace g by g/6
To determine whether the time period of a pendulum will remain the same, increase, or decrease on the surface of the moon, we need to understand the factors that influence the time period of a pendulum.
The time period of a pendulum is given by the formula:
T = 2π√(L/g)
Where T is the time period, L is the length of the pendulum, and g is the acceleration due to gravity.
In this case, we are comparing the acceleration due to gravity on the surface of the moon (1/6th of that on Earth) to this pendulum known as a second's pendulum. A seconds pendulum is a pendulum with a length such that its time period is exactly two seconds.
Let's assume the length of the seconds pendulum is L.
On the surface of the moon, the acceleration due to gravity is 1/6th of that on Earth. So, g_moon = (1/6)g_earth.
Substituting these values into the time period formula for the moon, we have:
T_moon = 2π√(L/g_moon)
= 2π√(L/((1/6)g_earth))
= 2π√(6L/g_earth)
Now, let's compare this with the time period of the seconds pendulum on Earth:
T_earth = 2π√(L/g_earth)
Comparing these two equations, we can see that T_moon = √6 * T_earth.
So, the time period of the seconds pendulum on the surface of the moon will increase compared to that on Earth. It will be approximately √6 times longer.
Therefore, the answer is that the time period of the pendulum will increase on the surface of the moon due to the lower acceleration due to gravity.