Question

the period of oscillation of a simple pendulum is 1.2 sec.in a place where g= 9.8m/s . how long is the bob below the rigid point?

Answer 1

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}$

$\green{T = 2π \sqrt{\frac{l}{g} }} \\ \pink{ \implies \: l \: = \frac{ {T }^{2} \times g }{4 {\pi}^{2} } } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \red{ \frac{ {(1.2)}^{2} \times 9.8}{4 \times {\pi}^{2} } \blue { = 0.357 \: \: m} }$

Answer 2

Answer:

Given: Period = T = 1.2 s, g = 9.8m/s2.

∴  l = 0.36 m

Explanation:

To Find: Length of pendulum = l =?