A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find the length of a seconds pendulum at a place where $g = π^{2} m s^{- 2} .$

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Solution

$T = 2 s e c$

$T = 2 π \sqrt{(\frac{l}{g})}$

$\Rightarrow 2 = 2 π \sqrt{(\frac{l}{10})}$

$\Rightarrow \frac{1}{10} = \frac{1}{π^{2}}$

$\Rightarrow l = 1 m (H H_{π^{2}} \approx 10)$

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A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find the length of a seconds pendulum at a place where g=π2 m s−2.

T=2 sec T=2π √(lg) ⇒ 2=2π √(l10) ⇒ 110=1π2 ⇒ l=1 m (HHπ2≈ 10)

A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find the length of a seconds pendulum at a place where $g = π^{2} m s^{- 2} .$

$T = 2 s e c$

$T = 2 π \sqrt{(\frac{l}{g})}$

$\Rightarrow 2 = 2 π \sqrt{(\frac{l}{10})}$

$\Rightarrow \frac{1}{10} = \frac{1}{π^{2}}$

$\Rightarrow l = 1 m (H H_{π^{2}} \approx 10)$