## University Physics with Modern Physics (14th Edition)

(a) We can find the period of the oscillation. $T = 2\pi~\sqrt{\frac{L}{g}}$ $T = 2\pi~\sqrt{\frac{6.50~m}{9.80~m/s^2}}$ $T = 5.117~s$ The climber reaches the lowest point after completing one-fourth of a cycle. We can find the time to reach the lowest point. $t = \frac{T}{4} = \frac{5.117~s}{4} = 1.28~s$ The time to reach the lowest point is 1.28 seconds. (b) The climber reaches the lowest point for the second time after completing three-fourths of a cycle. We can find the time to reach the lowest point for the second time. $t = \frac{3T}{4} = \frac{(3)(5.117~s)}{4} = 3.84~s$ The time to reach the lowest point for the second time is 3.84 seconds.