The acceleration due to gravity on the moon is 1.62 m/s2. What is the length of a pendulum whose period on the moon matches the period of a 4.00-m-long pendulum on earth?
To find the length of a pendulum that matches the period of a 4.00-m long pendulum on earth, we need to use the formula for the period of a pendulum:
T = 2π√(L/g)
Where:
T is the period of the pendulum
L is the length of the pendulum
g is the acceleration due to gravity
In this case, we are given the period of the 4.00-m pendulum on earth and the acceleration due to gravity on the moon. We can set up an equation and solve for the length of the pendulum on the moon.
First, let's find the period of the 4.00-m pendulum on earth. We know that T_earth = T_moon (since we want them to match). Let's say T_earth = T_moon = T.
Using the formula for the period of a pendulum, we can write the equation for the earth's pendulum:
T = 2π√(L_earth/g_earth)
Substituting the given values (L_earth = 4.00 m and g_earth = 9.81 m/s^2), we have:
T = 2π√(4.00/9.81)
Now, let's write the equation for the moon's pendulum:
T = 2π√(L_moon/g_moon)
Substituting the given values (g_moon = 1.62 m/s^2), we have:
T = 2π√(L_moon/1.62)
Since T_earth = T_moon, we can set these two equations equal to each other:
2π√(4.00/9.81) = 2π√(L_moon/1.62)
Now, we can solve for L_moon:
√(4.00/9.81) = √(L_moon/1.62)
Squaring both sides:
(4.00/9.81) = L_moon/1.62
Now, let's isolate L_moon:
L_moon = (4.00/9.81) * 1.62
Calculating this expression, we find:
L_moon ≈ 0.658 m
Therefore, the length of the pendulum on the moon whose period matches that of a 4.00-m long pendulum on earth is approximately 0.658 meters.