Question

The period T of a simple pendulum with small oscillations is calculated from the formula T=2pi sqrt(L/g) where L is the length of the pendulum and g is the acceleration due to gravity. suppose that measured values of L and g have errors and are corrected with new values where L is increased from 4m to 4.5m and g is increased from 9 m/s2 to 9.8 m/s2. Use differentials to estimate the change in the period. Does the period increase or decrease?

Answer 1

The step by step solution to the above problem is provided in the attached image file

period T of with small oscillations is given lay The Simble pandulam O T=27 Thon .Cds) dT 27 g2 2지 (d)L dT = g 2 2 2지 92 L 2 2772 L 2 7 272 P L P 4 2L 2 T d L 2 7 2 T d L 2 L 2T

& = g m L 4 m Then 4 L 27 T 2 7 4 71 1 (3) from 2 T 27 3 Now, Linreaoes 4mt o 4-5m C4-5-4 So d D 5 m increonen from 9to 9.8 m 5/u (6-8.6) Bp as = 0.8 m/s Now Substitwting +hese valns in Ramn We get O 5 O.8 2 4 Co.125 O 0888 2 dT Co.D362 dT (oo18) 20.618 - (o-0181) 4 (o.02413) T Sec 3 So, +he be riod increooen dT Co 0 2413) See