InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A simple pendulum of length 40 cm oscillates with an angular amplitude of 0.04 rad. Find a. the time period b. the linear amplitude of the bob, c. The speed of the bob when the strig makes 0.02 rad with the vertical and d. the angular acceleration when the bob is in moemntary rest. Take `g=10 ms^-2`. |
|
Answer» a. The angular frequency is `omega = sqrt((g)/(l)) = sqrt((10 m//s^(2))/(0.4 m)) = 5 s^(-1)` The time period is `(2 pi)/(omega) = (2 pi)/(5 s^(-1)) = 1.26 s` b. Linear amplitude `= 40 cm xx 0.4 = 1.6 cm` c. Angular speed `theta = omega sqrt(theta_(0)^(2) - theta^(2))` `theta = (5 s^(-1)) sqrt ((0.04)^(2) - (0.02)^(2)) rad = 0.17 rad//s` where speed of the bob at this instant `V = theta A = (40 cm) xx (0.17 s)` `= 6.8 cm//s` d. At momentary rest , the bob is in exterme position. Thus , the angular acceleration `|alpha| = omega^(2) theta`. `|alpha| = (0.04 rad) (25 s^(2)) = 1 rad//s^(2)` |
|