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 4.50-m-long pendulum on earth
For the same period, sqrt(l/g) must be the same. With 1/6 the value of g on the moon,
compared to Earth, you will need 1/6 the pendulum length to match the periods.
To find the length of a pendulum on the moon that has a period matching the period of a 4.50-m-long pendulum on Earth, we can use the formula for the period of a simple pendulum:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
Since we want to find the length of the pendulum on the moon (L_moon), we can rearrange the formula as follows:
T_earth = 2π√(4.50/g_earth)
T_moon = 2π√(L_moon/g_moon)
Since the periods are the same, T_earth = T_moon. We can substitute the values we know into the equation:
2π√(4.50/g_earth) = 2π√(L_moon/g_moon)
Cancel out the 2π on both sides:
√(4.50/g_earth) = √(L_moon/g_moon)
Now we can isolate L_moon by squaring both sides of the equation:
4.50/g_earth = L_moon/g_moon
Now, substitute the values for g_earth and g_moon:
4.50/9.81 = L_moon/1.62
Cross-multiply to solve for L_moon:
L_moon = (4.50/9.81) * 1.62
Calculating this gives us:
L_moon ≈ 0.744 meters
Therefore, the length of the pendulum on the moon that has a period matching the period of a 4.50-m-long pendulum on Earth is approximately 0.744 meters.