Question

if the length of a second pendulum increased by 1% how many seconds will it lose or gain in a day

Answer 1

Pendulum period in seconds
T ≈ 2π√(L/g)
L is length of pendulum in meters
g is gravitational acceleration = 9.8 m/s²

there are 86400 seconds in a day.

dT/dL = (1/2)2π√(1/g)√1/L) = π√(1/Lg)
dT = π/√(gL) dL

dL/L = 0.01 (ie, 1%)
dL = 0.01L
dT = π/√(gL) (0.01)L
dT = (0.01)(π/√g) L/√L
dT = (0.01)(π/√g)√L
dT = (0.01)(π√(L/g))
dT = (0.01)(1/2)(2π√(L/g))
dT = (0.005)T
dT = (0.005)(86400) = 432 seconds

___________2ND METHOD______________

By T = 2π√[L/g] ----------------(i)

=>T' = 2π√[1.01L/g]------------(ii)

=>T'/T = √[1.01]

=>T' = T x 1.004987562112089027021926491276

=>T' = 2 x 1.004987562112089027021926491276 {T = 2 sec, for a second pendulum}

=>T' = 2.0099751242241780540438529825519 sec

=>Lose = 2.0099751242241780540438529825519 - 2 = 0.0099751242241780540438529825519 sec

=>Lose/second = 0.00498756211208902702192649127596 sec
=>Loose in 24 hrs = 0.00498756211208902702192649127596 x 24 x 60 x 60
= 431.93 sec