Question

A simple pendulum executing SHM with a period of 6 s between two extreme positions B and C about a point O. If the length of the arc BC is 10 cm, how long will the pendulum take the move from position C to a position D towards O exactly midway between C and O?

• A. 0.5 s
• B. 1 s
• C. 1.5 s
• D. 3 s

Answer 1

Answer: (B)

1 s

Answer 2

O is the mean position and B and C are extreme positions

Given BC =10 cm

Since B and C are the extreme position therefore amplitude of the SHM oscillation is

D is the midway between C and O.

$\therefore \mathrm { DC } = \frac { 5 } { 2 } \mathrm {~cm}$

Since time is noted from extreme position hence displacement x at any time t is

$x = a \cos \omega t$

For displacement CD, let be the time taken. Then

$\frac { 5 } { 2 } = 5 \cos \omega t _ { 1 }$

$\frac { 1 } { 2 } = \cos \omega t _ { 1 }$ or $\cos \frac { \pi } { 3 } = \cos \omega t _ { 1 }$ or $t _ { 1 } = \frac { \pi } { 3 \omega }$

$( \because \mathrm { T } = 6 \mathrm {~s}$ Given $)$

$\therefore \mathrm { t } _ { 1 } = 1 \mathrm {~s}$