If the clock runs slow and loses 15 s per day, how should you adjust the length of the pendulum?

L = 1 m, ΔL = 0.0074 m

Explanation:

A clock is a simple pendulum with angular velocity

w = √ g / L

Angular velocity is related to frequency and period.

w = 2π f = 2π / T

We replace

2π / T = √ g / L

T = 2π √L / g

We will use the value of g = 9.8 m / s², the initial length of the pendulum, in general it is 1 m (L = 1m)

With this length the average time period is

T = 2π √1 / 9.8

T = 2.0 s

They indicate that the error accumulated in a day is 15 s, let's use a rule of proportions to find the error is a swing

t = 1 day (24h / 1day) (3600s / 1h) = 86400 s

e = Δt = 15 (2/86400) = 3.5 104 s

The time the clock measures is

T ' = To - e

T' = 2.0 - 0.00035

T' = 1.99965 s

Let's look for the length of the pendulum to challenge time (t ')

L' = T'² g / 4π²

L' = 1.99965 2 9.8 / 4π²

L ' = 0.9926 m

Therefore the amount that should adjust the length is

ΔL = L - L'

ΔL = 1.00 - 0.9926

ΔL = 0.0074 m