1.

A pendulum clock is accurate on the Earth's surface. How slow will it be, if it is lifted to the hundredth floor of a sky-scraper? The height of a storey is 3 m.

Answer»


Solution :The period of oscillations of the pendulum at the surface of the Earth is `T_(0)=2pisqrt(l//g_(0))`. At an ALTITUDE h above the surface of the Earth it is `T=2pisqrt(l//g)`. We have `(DELTAT)/(T_(0))=(tau)/(tau_(0))`, where `tau_(0)=8.64xx10^(4)s` is the duration of a complete day and `tau` is the lag of the clock. HENCE
`tau=(tau_(0)DeltaT)/(T_(0))=((T)/(T_(0))-1)tau_(0)(SQRT(g_(0)//g-1))`
The acceleration due to gravity at the Earth.s surface is `g_(0)=(gammaM)/(R^(2))` while at an altitude h it is g `(gammaM)/((R+h)^(2))` where R is the radius of the Earth. After some simple transformations we obtain
`tau_(0)h//R`


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