If the length of second’s pendulum is increased by 2%, how many seconds will it lose per day?

The time period (T) of a simple pendulum is given by the formula T = 2π√(L/g), where L is the length and g is the acceleration due to gravity [t4, t5]. For small changes, the fractional change in the time period is approximately half the fractional change in length (ΔT/T ≈ 1/2 * ΔL/L) [t1, t9]. If the length increases by 2%, the time period increases by approximately 1% (1/2 * 2%). A second's pendulum has a standard period of 2 seconds, meaning it completes 43,200 oscillations in a day (86,400 seconds). With a 1% increase in the period, the pendulum takes 1% more time for each oscillation, causing it to lag or 'lose' time. The total time lost per day is calculated as 1% of the total seconds in a day: 0.01 * 86,400 = 864 seconds [t3].