## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

We can find an expression for the original period. $T = 2\pi~\sqrt{\frac{L}{g}} = 4.00~s$ (a) Since the period does not depend on the mass, the period is still 4.00 seconds. (b) $T' = 2\pi~\sqrt{\frac{2L}{g}}$ $T' = \sqrt{2}\times ~2\pi~\sqrt{\frac{L}{g}}$ $T' = \sqrt{2}~T$ $T' = \sqrt{2}~(4.00~s)$ $T' = 5.66~s$ (c) $T' = 2\pi~\sqrt{\frac{L/2}{g}}$ $T' = \frac{1}{\sqrt{2}}\times ~2\pi~\sqrt{\frac{L}{g}}$ $T' = \frac{1}{\sqrt{2}}~T$ $T' = \frac{1}{\sqrt{2}}~(4.00~s)$ $T' = 2.83~s$ (d) Since the period does not depend on the amplitude, the period is still 4.00 seconds.