Answer :
Final answer:
The question pertains to calculating the new period of oscillation for a skydiver-parachute system in simple harmonic motion when a second skydiver is added to the system. The period of oscillation increases with the additional mass, and this is determined by using the formula for a simple harmonic oscillator, taking into account the unchanged spring constant.
Explanation:
When considering the oscillation period of a skydiver hanging from a parachute, we're dealing with a problem from the realm of Physics, specifically related to simple harmonic motion. In this scenario, the initial period of the oscillation for the first skydiver, who has a mass of 94 kg, is 1.5 seconds. The question seeks to determine the new period of oscillation when another skydiver, with a mass of 69 kg, hangs from the first skydiver. This system's period can be found by analyzing the formula for the period of a simple harmonic oscillator, T = 2π√(m/k), where 'T' is the period, 'm' is the total mass of the system, and 'k' is the spring constant or the force constant of the parachute harness system.
Assuming the parachute's harness behaves like a spring, the spring constant 'k' remains unchanged when the second skydiver hangs on the first. Since the period of a simple harmonic oscillator is directly proportional to the square root of the mass, the new total mass of the system (the sum of both skydivers' masses) will affect the period. To find the new period, T', we use the ratio of the periods squared equal to the ratio of the masses, (T'/T)² = (m + m')/m, where 'm' is the original mass and 'm' is the second skydiver's mass. Solving for T', we get a new period that depends on the increased total mass of the system, showing that the period will increase when additional mass is added.