A pendulum has a period of 0.70 s on Earth. What is its period on a planet where the acceleration of gravity is about 0.31 that on Earth?
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T1=2π•sqrt(L/g)
T2=2π•sqrt(L/0.31g)
T2=T1/sqrt(0.31)=0.7/0.557=1.26 s.
To find the period of a pendulum, we use the equation:
T = 2π √(L / g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
In this case, we are given that the period of the pendulum on Earth (T₁) is 0.70 s.
We need to find the period on the other planet (T₂), where the acceleration of gravity is 0.31 times that on Earth.
Since the length of the pendulum is not given, we can assume that it remains the same on both Earth and the other planet. Therefore, we can ignore the length (L) in our calculations.
Using the equation for the period of a pendulum, we can write the following ratio:
T₂ / T₁ = √(g₂ / g₁)
where g₂ is the acceleration due to gravity on the other planet and g₁ is the acceleration due to gravity on Earth.
We are given that g₂ = 0.31g₁.
Plugging this value into the equation, we have:
T₂ / T₁ = √(0.31g₁ / g₁)
Simplifying the equation:
T₂ / T₁ = √0.31
To find T₂, we can rearrange the equation:
T₂ = T₁ * √0.31
Substituting the given value for T₁:
T₂ = 0.70 s * √0.31
Calculating this expression:
T₂ = 0.70 s * 0.556 ≈ 0.3892 s
Therefore, the period of the pendulum on the planet with an acceleration of gravity about 0.31 times that on Earth is approximately 0.3892 seconds.