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.
An inspection of the equation for the period shows that the period Increases as g decreases.
Therefore, the period is Longer on the moon.
To determine whether the time period of a pendulum will remain the same, increase, or decrease on the surface of the moon where the acceleration due to gravity is 1/6th of that on Earth, we need to understand the factors that affect the time period of a pendulum.
The time period of a simple 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 know that the acceleration due to gravity on the moon is 1/6th of that on Earth. Therefore, the value of g on the moon would be (1/6) * g, where g is the acceleration due to gravity on Earth.
Let's consider a second's pendulum, which is a pendulum with a length equal to one second of time period on Earth. So, L = 1 second.
Now, let's compare the time periods on Earth and the moon.
On Earth:
T_earth = 2π √(1 / g_earth)
On the moon:
T_moon = 2π √(1 / [(1/6) * g_earth])
To simplify this expression, we can cancel out the π and rearrange the terms:
T_moon = T_earth √(g_earth / (1/6) * g_earth)
= T_earth √(6)
As we can see, the time period on the moon is equal to the time period on Earth multiplied by the square root of 6. Therefore, the time period of the pendulum on the moon will increase compared to its time period on Earth.
In conclusion, the time period of a second's pendulum on the surface of the moon, where the acceleration due to gravity is 1/6th of that on Earth, will increase due to the lower gravitational acceleration on the moon.
period will be shorter...gravity on earth is stronger, and it pulls the pendulum down faster.
period=2PIsqrt(Length/g)