if a simple pendulum is driven clock gains 10 seconds per day, what fractional change in the pendulum length must be made for it to keep perfect time
see this previous response for a slightly different error. https://www.jiskha.com/display.cgi?id=1510151111
the period of a simple pendulum is proportional to the square root of its length
P = k √L
the clock is gaining time , so the period is too short
seconds in a day ... 24 * 60 * 60 = 86400
... the error is one part in 8640 ... about a hundredth of a percent
P1 / √L1 = P2 / √L2 ... L2 = L1 (P2 / P1)^2 ... L2 = L1 {1 / [1 - (1 / 8640)]}^2
To find the fractional change in the pendulum length that must be made for it to keep perfect time, we can use the concept of the period of the simple pendulum.
The period of a simple pendulum is the time it takes for the pendulum to complete one full swing back and forth. The period is given by the formula:
T = 2π√(L/g)
Where:
T is the period (in seconds)
L is the length of the pendulum (in meters)
g is the acceleration due to gravity (approximately 9.8 m/s² on Earth)
In this case, the pendulum is driven, so its period will be slightly different from the ideal period given by the formula above.
If the clock gains 10 seconds per day, it means that the driven pendulum takes 10 seconds more than the ideal period to complete one full swing. Let's call this new period T_driven.
T_driven = T + 10
To keep perfect time, the driven pendulum should have the same period as the ideal pendulum. So we need to find the fractional change in the pendulum length that compensates for the 10-second difference.
We can rearrange the formula for the period as follows:
T = 2π√(L/g)
T² = (4π²L)/g
L = (g/4π²)T²
Where:
T is the ideal period
L is the ideal length
Now, let's substitute the expressions for T and L in terms of T_driven into the equation above:
L_driven = (g/4π²)T_driven²
L_driven = (g/4π²)(T + 10)²
To find the fractional change in the pendulum length, we calculate the difference between the driven length (L_driven) and the ideal length (L), and divide it by the ideal length:
ΔL/L = (L_driven - L)/L
Substituting the expressions for L and L_driven into the equation above:
ΔL/L = [(g/4π²)(T + 10)² - (g/4π²)T²]/[(g/4π²)T²]
Simplifying the expression:
ΔL/L = [(T + 10)² - T²]/T²
ΔL/L = [(T² + 20T + 100) - T²]/T²
ΔL/L = (20T + 100)/T²
Since the period T is related to the length of the pendulum through the equation T = 2π√(L/g), we can substitute this expression into the equation for ΔL/L:
ΔL/L = (20(2π√(L/g)) + 100)/(2π√(L/g))²
Simplifying further:
ΔL/L = (40π√(L/g) + 100)/(4π²(L/g))
Now, we can cancel out the common factors of π, and g:
ΔL/L = (10√(L/g) + 25)/(π(L/g))
Therefore, to keep perfect time, the fractional change in the pendulum length must be (10√(L/g) + 25)/(π(L/g)).