A pendulum clock keeps perfect time at a location where the acceleration due to gravity is exactly 9.8 m s⁻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. (9.78 m s⁻²)



Given:

Value of acceleration due to gravity = g = 9.8 m s⁻²
 = F  = 0.4 N
Time lost in a day  = 80.0 s 

Time lost in a second = (80.0/3600) s = 0.000926 s 

To Find:


Value of g at a new location = g' = ?


Solution:


Let the Time Period of a pendulum which keeps perfect time = T = 1 s

T = 2π `\sqrt frac{L}{g}` = 1 s -----------------(1)


The loss of 80 s per day means that the pendulum swings slower and the time period will be greater than 1s of the previous location. According to the given data

T' = T + time lost per seconds = 1 s + 0.000926 s = 1.000926 s

so the Time period in a new place 

T' = 2π `\sqrt frac{L}{g'}` = 1.000926 s -------------(2)

In order to get the value of g' we will divide equation (2) by eqn (1) 

`\frac {T'}{T}` = `\frac {2π sqrt frac{L}{g'}}{2π sqrt frac{L}{g}}`

or

`\frac {T'}{T}` = `\sqrt frac {g}{g'}`

or

g' = `\(frac {T}{T'})^2` g

putting values

g' = `\(frac {1 s}{1.000926 s})^2` ｘ 9.8 m s⁻²

g' = `\(0.999)^2` ｘ 9.8 m s⁻²

g' = 0.998 ｘ 9.8 m s⁻²

g' = 9.78 m s⁻² ----------------Ans

Note: Keeping in view that the length of the pendulum was the same at both locations. So, to get the perfect time at this new location where g' = 9.78 m s⁻² we must calculate - cum - adjust the length of the pendulum.

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