A pendulum clock was moved from a location where g=9.8152 m/s^2 to another location where g=9.7959 m/s^2. During the move, the length of the clock's pendulum did not change; nevertheless, the clock lost accuracy. Assuming the clock was perfectly accurate at its previous location, how many seconds a day does it lose at the new location?
Answer in units of seconds.
To determine the number of seconds the clock loses in a day, we need to calculate the difference in gravitational acceleration (g) between the two locations.
Let's define the time the clock takes for one complete swing as T. According to the law of simple pendulum motion, the time period of a pendulum is given by the formula:
T = 2π√(L/g)
Where L is the length of the pendulum and g is the acceleration due to gravity.
Since the length of the pendulum does not change, we can use the formula to compare the time periods at the two locations:
T1 = 2π√(L/g1) (Previous location)
T2 = 2π√(L/g2) (New location)
To find the difference in time, we subtract T2 from T1:
ΔT = T1 - T2
= 2π√(L/g1) - 2π√(L/g2)
= 2πL(√(1/g1) - √(1/g2))
Now, we are given the values of g1 = 9.8152 m/s^2 and g2 = 9.7959 m/s^2. Substituting these values into the equation, we can calculate the difference in time:
ΔT = 2πL(√(1/9.8152) - √(1/9.7959))
Once we have calculated ΔT, we know that the clock loses this amount of time per swing. Since there are 24 hours in a day, and 60 minutes in an hour, and 60 seconds in a minute, we can convert this value to seconds lost per day:
Seconds lost per day = ΔT * (24 * 60 * 60)
Evaluate the equation to find the answer in units of seconds.