Applications Of Dimensional Analysis

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Assuming the time period of a simple pendulum can depend only on its length, its mass and the acceleration due to gravity, which relation correctly emerges from dimensional analysis?

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Dimensional analysis can reveal the form of a physical relationship by demanding that both sides share the same dimensions. Assume T = k l^a m^b g^c, where k is a dimensionless constant. Writing dimensions, [T] equals [L]^a [M]^b [L T^-2]^c. Collecting powers gives mass: b = 0; length: a + c = 0; time: -2c = 1. Solving yields c = -1/2, a = 1/2, and b = 0, so T ∝ l^(1/2) g^(-1/2) = √(l/g). The result confirms the well-known independence of the pendulum period from mass. The choice √(l g) has the wrong sign on the gravity exponent and gives dimension [T^-1], not time. The choice l/g produces dimension [T^2], the square of a time, so it is wrong. The choice √(m l/g) wrongly retains a mass dependence that the analysis explicitly forbids since b = 0. This is the classic NCERT illustration of deducing relations dimensionally. A check confirms it: √(l/g) has dimensions √([L]/[L T^-2]) = √[T^2] = [T], exactly a time.

Applications Of Dimensional Analysis

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