Question #d58fd

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Question #d58fd

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Answer 1 · 2016-05-14T15:10:42

$2 \sqrt{2} s$ $2sqrt2s$

Explanation:

For small angle of deviation $θ$ $theta$ from the mean position the time period $T$ $T$ of a pendulum is given as

$T = 2 π \sqrt{\frac{L}{g}}$ $T=2pisqrt(L/g)$ .......(1)
where $L$ $L$ is the length of the pendulum and $g$ $g$ is the local gravity.

We know that seconds pendulum is a pendulum which has its period of two seconds. Its each swing takes one second, $T = 2 s$ $T=2s$
Also local acceleration due to gravity $g = G \frac{M_{e}}{R_{e}^{2}}$ $g=G (M_e)/R_e^2$ .
$G$ $G$ is constant $= 6.67408 \times 10^{- 11} m^{3} k g^{- 1} s^{- 2}$ $=6.67408 xx 10^-11 m^3 kg^-1 s^-2$ . $M_{e}$ $M_e$ is mass and $R_{e}$ $R_e$ is radius of earth respectively.

Now the acceleration due gravity of the planet is

$g_{p} = G \frac{M_{p}}{R_{p}^{2}}$ $g_p=G (M_p)/R_p^2$

Inserting given quantities we get

$g_{p} = G \frac{2 M_{e}}{{(2 R_{e})}_{e}^{2}}$ $g_p=G (2M_e)/(2R_e)^2$
$\Rightarrow g_{p} = G \frac{M_{e}}{2 {(R_{e})}_{e}^{2}}$ $=>g_p=G (M_e)/(2(R_e)^2)$ ,
in terms of $g$ $g$
$g_{p} = \frac{g}{2}$ $g_p=g/2$

Inserting in (1)

$T_{p} = 2 π \sqrt{\frac{L}{g_{p}}}$ $T_p=2pisqrt(L/g_p)$ ........(2)

in terms of $g$ $g$

$T_{p} = 2 π \sqrt{2 \frac{L}{g}}$ $T_p=2pisqrt(2L/g)$

Using (1)

$T_{p} = \sqrt{2} T$ $T_p=sqrt2T$
$\Rightarrow T_{p} = 2 \sqrt{2} s$ $=>T_p=2sqrt2s$

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Question #d58fd

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