A simple pendulum of length $L_{1}$ has a time period ' $T_{1}$ ' and another pendulum of length ' $L_{2}$ ' has time period ' $T_{2}$ '. Then the time period of pendulum of length ( $L_{1} - L_{2}$ ) is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$T_{1} + T_{2}$

$\sqrt{T_{1} + T_{2}}$

$\sqrt{T_{1} T_{2}}$

$\sqrt{{T_{1}}^{2} {-T}_{2}^{2}}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$t i m e p e r i o d o f s i m p l e p e n d u l u m i s T_{1} = 2π \sqrt{\frac{l_{1}}{g}} ⇒ {T_{1}}^{2} α l_{1} - - - (1) o b t a i n e d o n s q u a r i n g b o t h s i d e s a n d 2 π, g are constants T_{2} = 2π \sqrt{\frac{l_{2}}{g}} ⇒ {T_{2}}^{2} α l_{2} - - - (2) l i s l e n g t h o f p e n d u l u m, g i s a c c e l e r a t i o n d u e t o g r a v i t y T = 2π \sqrt{\frac{l_{1} - l_{2}}{g}}, o n s q u a r i n g b o t h s i d e s, a n d 2 π, g are constants \Rightarrow T^{2} α l_{1} - l_{2} s u b s t i t u t i n g v a l u e s f r o m e q u a t i o n s (1), (2) T^{2} α {T_{1}}^{2} - T_{2}^{2} T α \sqrt{{T_{1}}^{2} - {T_{2}}^{2}}$