Answer
(a) The time to reach the lowest point is 1.28 seconds.
(b) The time to reach the lowest point for the second time is 3.84 seconds.
Work Step by Step
(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.
