Problem 2: A pendulum clock keeps perfect time at a location where the acceleration due to gravity is exactly 9.8 ms-2. When the clock is moved to a higher altitude, it loses 80.0 s per day. Find the value of g at this new location?

Solution

Explanation: The general equation for the motion of a pendulum is given by
Therefore, to find g at any location, we must know about the time period T and length l of the pendulum at that location. We can find the time period at the new location from the given data. If the clock loses 80 s in 24 hours, then the time period in the new location can be calculated. It should be remembered that if the clock keeps perfect time in the first location then its time period is exactly 1 s.

Now, time loss in 24 hrs (= 24 × 60 × 60 = 86400 s) = 80 s

Time loss in 1 s = 80/86400 = 0.000926 s

Time period at the new location = 1 s + 0.000926 s = 1.000926 s = T_{2} … (1)

Method 1: For the first location, put T = 1 s and g = 9.8 ms^{-2}, the equation of pendulum becomes,

Therefore,

Let the value of gravitational acceleration at the new location = g_{2}

Equation of pendulum at the new location

Put the value of T_{2} from result (1) and l from (2),

Method 2:

Hint: Divide equation (A) by equation (B) and simplify to get the value of g_{2}.

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