Use T = 2π√(L/g_eff) and Decide What g_eff Is
A simple pendulum has T = 2π sqrt(L / g_eff). Questions change either the length or the effective gravity, so the first job is deciding what g_eff is in the situation described.
Simple Pendulum and Effective Gravity is the central idea in this question pattern. Identify the condition in the stem before applying the method below.
Length and effective gravity
T is proportional to sqrt(L): increasing the length by 44% multiplies T by sqrt(1.44) = 1.2, so a 60 s pendulum becomes 72 s. For a pendulum hung in a vehicle sliding freely (without friction) down an incline of angle α, the effective gravity in the vehicle's frame is g cos α, so T = 2π sqrt(L / (g cos α)). In an artificial satellite the effective gravity is zero, so the pendulum does not swing and a pendulum clock cannot be used there.
A pendulum as long as the Earth
With L very large (infinite) it becomes 2π sqrt(R / g) = 84.6 min, which is also the period of a satellite just above the surface and of a ball oscillating in a tunnel through the Earth's centre: all three are 84.6 minutes.
OlympIQ exam lens
- ✓T = 2π sqrt(L / g_eff); T is proportional to sqrt(L)
- ✓Frictionless incline of angle α: g_eff = g cos α
- ✓Free fall or orbit: g_eff = 0, so no oscillation
- ✓Pendulum of length comparable to R: 1/L + 1/R replaces 1/L
- ✓84.6 min is the period for L → ∞, a surface satellite and a tunnel oscillation
Do not use g in the pendulum formula when the support is accelerating. Whenever the point of suspension accelerates, replace g by the effective gravity in that frame, which can be larger, smaller or even zero.
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