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A 90.0 -kg skydiver hanging from a parachute bounces up and down with a period of 1.50 s. What is the new period of oscillation when a second skydiver, whose mass is 60.0 $\mathrm{kg}$ hangs from the legs of the first, as seen in Figure 16.47Figure 16.47 The oscillations of one skydiver are about to be affectedby a second skydiver. (credit: U.S. Army, www. army.mil)As usual, the acceleration due to gravity in these problems is taken to be $g=9.80 \mathrm{m} / \mathrm{s}^{2},$ unless otherwise specified.
$T_{2}=1.936 \mathrm{s}$
Physics 101 Mechanics
Physics 103
Chapter 16
Oscillatory Motion and Waves
Periodic Motion
Wave Optics
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So to solve this question, we're going to make use off formula T equal to to pi square root k divided. But sorry. M divided by K So in this supposed part of the question states that 90 cages skydiver is hanging on his period oscillation frequency is 1.5. So we are given T and a mass so we can use this formula to find the value of K. O. K. Is equal to to buy Bye t squared times to mess. So a substitute on the values we know cheese 1.5 squared times the mass, which is 90 kilograms. So if you saw this, you find the answer. Here is 1005 18 new turn per meter. Now let's substitute that value to find the new frequency, which is two pi times again mass divided by K. But now we know mass and the case in this case mass is 90 plus 60 because two divers are handling suits 1 50 divided by 1000 Fire 80. So if you saw this, you get the value US one point nine 36 seconds. Thank you
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