Question

6. Find the length of a simple pendulum whose time period at a place is 2 second and g = 9.8 ms.​

Answer 1

Explanation:

$= > 2 = 2\pi \sqrt{ \frac{l}{9.8} } \\ = > 4 = 4 \times {\pi}^{2} \frac{l}{9.8} \\ = > \frac{9.8}{ {\pi}^{2} } = l \\ = > l = 0.98 \: m$

(π² is approximately equal to 10)

Answer 2

Given :

▪ Time period of pendulum = 2s

▪ Acc. due to gravity = 9.8m/s²

To Find :

▪ Length of the simple pendulum.

Formula :

☞ Formula of time period of a simple pendulum in terms of length of pendulum and acceleration due to gravity is given by

✴ T = 2π × √(l/g)

where,

T denotes time period

l denotes length

g denotes gravitational acceleration

Calculation :

→ T = 2π × √(l/g)

→ 2 = 2π × √(l/9.8)

→ l/9.8 = 1/π²

→ l = 9.8/π²

→ l = 0.99 ≈ 1m

Additional information :

❇ Second pendulum is the simple pendulum, having a time period of 2 second. Its effective length is 99.992 (≈1m) on earth.