The periodic time T 1 of a simple pendulum is 4 s at a place, at a height R above the earth's surface. If the pendulum is brought at a height 3 2 R above the earth surface, the periodic time

Question

The periodic time T 1 ​ of a simple pendulum is 4 s at a place, at a height R above the earth's surface. If the pendulum is brought at a height 3 2 ​ R above the earth surface, the periodic time decreases by x T 1 ​ . The value of x is

TestHub · Accepted Answer

The time period of a simple pendulum is given by $T = 2 π \frac{l}{g}$ .

The acceleration due to gravity at a height $h$ from the surface of the earth is $g_{h} = \frac{GM}{( R + h ) ^{2}}$ .

Thus, $T \propto \frac{1}{g _{h}} \propto (R + h)$ .

At height $h_{1} = R$ , $T_{1} \propto (R + R) = 2 R$ .

At height $h_{2} = \frac{2}{3} R$ , $T_{2} \propto (R + \frac{2}{3} R) = \frac{5}{3} R$ .

Taking the ratio, we get $\frac{T _{2}}{T _{1}} = \frac{\frac{5}{3} R}{2 R} = \frac{5}{6}$ .

$T_{2} = \frac{5}{6} T_{1}$

The decrease in time period is $Δ T = T_{1} - T_{2} = T_{1} - \frac{5}{6} T_{1} = \frac{1}{6} T_{1}$ .

Comparing this with $x T_{1}$ , we get $x = \frac{1}{6}$ .

Answer: $\frac{1}{6}$

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