Question

A simple pendulum is vibrating with an angular amplitude of π/2. The value of a for which the resultant acceleration has a direction along the horizontal is

• A. $\frac { \pi } { 2 }$
• B. 180⁰
• C. $\cos ^ { - 1 } \left( \frac { 1 } { \sqrt { 3 } } \right)$
• D. $\cos ^ { - 1 } \left( \frac { 1 } { \sqrt { 2 } } \right)$

Answer 1

Answer: (C)

$\cos ^ { - 1 } \left( \frac { 1 } { \sqrt { 3 } } \right)$

Answer 2

Here a_r = $\frac { v ^ { 2 } } { 1 } = \frac { ( \sqrt { 2 g ( 1 \cos \alpha ) } ) ^ { 2 } } { 1 }$

= $\frac { 2 g l \cos \theta } { 1 }$ = 2g cos α

and tan α = $\frac { a _ { r } \sin 90 ^ { \circ } } { a _ { r } + a _ { r } \cos 90 ^ { \circ } } = \frac { 2 g \cos \alpha } { g \sin \alpha }$

( a_t = g sin α)

= $\frac { 2 } { \tan \alpha }$

i.e., tan²α = 2 or sec² α = 3

or cos α = $\frac { 1 } { \sqrt { 3 } }$

or α = cos^-1 $\left( \frac { 1 } { \sqrt { 3 } } \right)$