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Section 3.9– Number 21 Writing g for the acceleration due to gravity, the period, T, of a pendulum of length l is given by: l g T = 2π a) Show that if the length of the pendulum changes by ∆l , the change in the period, ∆T , is given by ∆T = T ∆l 2l Solution: First, we need to rewrite this formula "separating" the constants from the variables. 1 l2 T = 2π = g dT dl Finding 2π g 1 l2 , we get: dT 2π 1 − 21 π 1 = i l = dl 2 g g l Or dT = π 1 g i l dl This is equivalent to ∆T = π g 1 i l ∆l Recall that T = 1 T g = 2π l 2 2π g 1 l2 . Multiplying both sides by g , we get: Solving for g , we get: 1 2π l 2 g = T Substituting this into the ∆T formula, we get: ∆T = π 2π l T 1 1 2 i Thus ∆T = l 1 2 ∆l = π i T 2π l 1 1 2 i 1 ∆l l2 T ∆l . 2l b) If the length of the pendulum increases by 2%, by what percent does the period change? Solution: We have that ∆l = 0.02. Substitute this into the ∆T formula. So ∆T = T T (0.02) = (0.01) 2l l Thus the period change is 1%.