A pendulum clock is accurate at a place where $g = 9.8 m / s^{2}$ . Find the value of g at another place where clock becomes slow by 24 seconds in a day (24 hrs).

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Solution

We know that $T = \sqrt{\frac{L}{g}}$ [where length is assumed to be constant]
$\frac{Δ T}{T} = \frac{1}{2} \frac{Δ g}{g}$
But $Δ T = 24 s$ (given) and $T = (24 \times 3600) s$
$\frac{Δ T}{T} = \frac{24}{24 \times 3600} = \frac{1}{3600}$
$\frac{Δ g}{g} = 2 \frac{Δ T}{T} = 2 \times \frac{1}{3600} = \frac{1}{1800}$
$Δ g = \frac{9.81}{1800} = + (9.81 - g^{'})$
$g^{'} = (9.81 - \frac{9.81}{1800}) = 9.8045 m / s^{2}$
$g^{'} = 9.8045 m / s^{2}$

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A pendulum clock is accurate at a place where g=9.8m/s2. Find the value of g at another place where clock becomes slow by 24 seconds in a day (24 hrs).

We know that T=√Lg [where length is assumed to be constant]ΔTT=12ΔggButΔT=24s(given) and T=(24×3600)sΔTT=2424×3600=13600Δgg=2ΔTT=2×13600=11800Δg=9.811800=+(9.81−g′)g′=(9.81−9.811800)=9.8045m/s2g′=9.8045m/s2

A pendulum clock is accurate at a place where $g = 9.8 m / s^{2}$ . Find the value of g at another place where clock becomes slow by 24 seconds in a day (24 hrs).