The bob of a swinging seconds pendulum (one whose time period is 2 s ) has a small speed v0 at its lowest point. Its height from this lowest point 2.25 s after passing through it is given by

Question

The bob of a swinging seconds pendulum (one whose time period is 2 s ) has a small speed v0 at its lowest point. Its height from this lowest point 2.25 s after passing through it is given by (A) (v0 2/2 g) (B) (v0 2/g) (C) (v0 2/4 g) (D) (9 v0 2/4 g)

Tardigrade Admin · Accepted Answer

T =2 sec. at t=T+T / 8 v=A ω cos ω t=(v0/√2) By Mechanical energy conservation: ((v0/√2))2=v02-2 g h h=(v02/4 g)

Q. The bob of a swinging seconds pendulum (one whose time period is $2 s$ ) has a small speed $v_{0}$ at its lowest point. Its height from this lowest point $2.25 s$ after passing through it is given by

$\frac{v _{0} ^{2}}{2 g}$

$\frac{v _{0} ^{2}}{g}$

$\frac{v _{0} ^{2}}{4 g}$

$\frac{9 v _{0} ^{2}}{4 g}$

$T = 2 sec$ .
at $t = T + T /8$
$v = A ω cos ω t = \frac{v _{0}}{2}$
By Mechanical energy conservation:
$(\frac{v _{0}}{2})^{2} = v_{0}^{2} - 2 g h$
$h = \frac{v _{0}^{2}}{4 g}$

Q. The bob of a swinging seconds pendulum (one whose time period is 2s ) has a small speed v0​ at its lowest point. Its height from this lowest point 2.25s after passing through it is given by

2gv0​2​

gv0​2​

4gv0​2​

4g9v0​2​