Question

Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period T and the length L of a simple pendulum with the equation:

T = 2π√(L/g),

where g is the constant acceleration of gravity at the pendulum’s location. If we measure g in centimeters per second squared, we measure L in centimeters and T in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to L. In symbols, with u being temperature and k the proportionality constant,

dL/du = kL.

Assuming this to be the case, show that the rate at which the period changes with respect to temperature is kT/2.

Accepted Answer

Start by differentiating the given equation for the period of the pendulum, T = 2π√(L/g), with respect to temperature u. This requires using the chain rule since L is a function of u. Apply the chain rule: dT/du = (dT/dL) * (dL/du). First, find dT/dL by differentiating T = 2π√(L/g) with respect to L. This gives dT/dL = π/√(Lg). Substitute the expression for dL/du from the problem, which is dL/du = kL, into the chain rule expression. This gives dT/du = (π/√(Lg)) * (kL). Simplify the expression for dT/du. Notice that L/√L simplifies to √L, so dT/du = πk√(L/g). Recognize that T = 2π√(L/g) from the original equation. Therefore, dT/du = (k/2) * T, which shows that the rate at which the period changes with respect to temperature is kT/2.

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Key Concepts

Differentiation

Chain Rule

Proportionality and Linear Differential Equations