Adjusting Pendulum Length to Correct Clock Error

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

Answer: 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’= 1.99965² 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

If the clock is losing time, should you increase or decrease the length of the pendulum to correct the error?

To correct the error of a clock that loses time, you should decrease the length of the pendulum. The calculation shows that an adjustment of 0.0074 meters shorter is needed to make the clock keep accurate time.