Question

Calculate length of seconds pendulum at a place where g= 9.78 m/s^2(meter/ second square).

Answer 1

Answer:

$\boxed{\sf Length \ of \ seconds \ pendulum \ (l) = 0.99 \ m \approx 1 \ m}$

Given:

Acceleration due to gravity (g) = 9.78 m/s²

To Find:

Length of seconds pendulum (l)

Explanation:

Time period (T) for one oscillation in seconds pendulum is 2 seconds.

$\sf \implies T = 2 \ s$

Formula:

$\boxed{ \bold{ \sf T = 2\pi \sqrt{ \frac{l}{g} } }}$

$\sf \implies l = \frac{T ^{2} g}{4 {\pi}^{2} }$

Substituting values of T & g in the equation:

$\sf \implies l = \frac{ {2}^{2} \times 9.78 }{4 \times3.14 ^{2} }$

$\sf \implies l = \frac{ \cancel{4} \times 9.78}{ \cancel{4} \times 9.8596}$

$\sf \implies l = \frac{ 9.78}{ 9.8596}$

$\sf \implies l = 0.99 \: m \approx 1 \: m$

$\therefore$

Length of seconds pendulum (l) = 0.99 m ≈ 1 m

Answer 2

Given ,

Acceleration due to gravity (a) = 9.8 m/s²

We know that , the time period simple pendulum is given by

$\boxed{ \tt{T = 2\pi \sqrt{ \frac{l}{g} } }}$

And the time period of second pendulum is 2 sec

Thus ,

$\mapsto \tt 2 = 2\pi \sqrt{ \frac{l}{9.8} }$

Squaring on both sides , we get

$\mapsto \tt \cancel4 = \cancel4 \times {(3.14)}^{2} \times \frac{l}{9.8}$

$\mapsto \tt l = \frac{9.8 }{ {(3.14)}^{2}}$

$\mapsto \tt l = \frac{9.8}{9.8596}$

$\mapsto \tt l = 1 \: m \: (approx)$

★ The length of second pendulum is 1 m