A 34-kg child on an 18-kg swing set swings back and forth through small angles.If the length of the very light supporting cables for the swing is $4.9 \mathrm {~m}$ how long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables.

Question

Quizplus · Accepted Answer

The period of a pendulum is given by $T = 2\pi\sqrt{\frac{l}{g}}$, where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity ($9.8 \mathrm{~m/s^2}$). Plugging in the given length of $4.9 \mathrm{~m}$, we get $T = 2\pi\sqrt{\frac{4.9}{9.8}} = 2\pi\sqrt{0.5} = 2\pi\sqrt{\frac{1}{2}} = 2\pi(\frac{1}{\sqrt{2}}) \approx 4.4 \mathrm{~s}$. The mass of the child and swing set does not affect the period of the swing for small angles, so the answer is 4.4 seconds.

A 34-Kg Child on an 18-Kg Swing Set Swings Back 4.9 m4.9 \mathrm {~m}4.9 m

A 34-Kg Child on an 18-Kg Swing Set Swings Back $4.9 \mathrm {~m}$