The time period of a physical pendulum about some pivot point is T . When we take another pivot point, opposite of the first one such that the centre of mass of the physical pendulum lies on the line joining these two pivot points, we obtain the same time period. If the two points are separated by a distance L , then the time period T is

Question

The time period of a physical pendulum about some pivot point is T . When we take another pivot point, opposite of the first one such that the centre of mass of the physical pendulum lies on the line joining these two pivot points, we obtain the same time period. If the two points are separated by a distance L , then the time period T is (A) 2π √(L/g) (B) π √(L/g) (C) 2π √(L/2 g) (D) π √(2 L/g)

Tardigrade Admin · Accepted Answer

T=2 π √(IC+m x2/m g x)=2 π √(IC+m(L-x)2/m g(L-x)) (IC/x)+mx=(IC/L - x)+m(.L-x.) IC((L - 2 x/x (. L - x .)))=m(.L-2x.) IC=mx (.L-x.) T=2π √(m x L - m x2 + m x2/m g x)=2π √(L/g)

$2 π \frac{L}{g}$

$π \frac{L}{g}$

$2 π \frac{L}{2 g}$

$π \frac{2 L}{g}$

$T = 2 π \frac{I _{C} + m x ^{2}}{m gx} = 2 π \frac{I _{C} + m ( L - x ) ^{2}}{m g ( L - x )}$
$\frac{I _{C}}{x} + m x = \frac{I _{C}}{L - x} + m (L - x)$
$I_{C} (\frac{L - 2 x}{x ( L - x )}) = m (L - 2 x)$ $I_{C} = m x (L - x)$
$T = 2 π \frac{m xL - m x ^{2} + m x ^{2}}{m gx} = 2 π \frac{L}{g}$

2πgL​​

πgL​​

2π2gL​​

πg2L​​

T=2πmgxIC​+mx2​​=2πmg(L−x)IC​+m(L−x)2​​ xIC​​+mx=L−xIC​​+m(L−x) IC​(x(L−x)L−2x​)=m(L−2x) IC​=mx(L−x) T=2πmgxmxL−mx2+mx2​​=2πgL​​

$2 π \frac{L}{g}$

$π \frac{L}{g}$

$2 π \frac{L}{2 g}$

$π \frac{2 L}{g}$