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Part A选择题部分:
Part B简答题部分
Solution to Problem 2:
From the photos one can measure that the height of the woman of 1.68 m corresponded to 35 mm so the conversion was 0.048 m/mm.
One could measure the change of height of center of gravity of the person from the position in the air, at the top of the jump to the bottom when the spring was fully compressed. It was 7 mm which corresponds to 0.34 m. One could also measure the compression of the spring x (how much shorter the bottom of the pogo stick rod looks) to be 5 mm or 0.24 m.
All gravitational potential energy is converted to the elastic potential energy of the spring
mgh = kx2/2
where m = 60 kg (mass of the pogo stick can be neglected or, as some students dis estimated to be of the order of 5kg), h = 0.34 m, x = 0.24 m, k is the elastic constant of the spring.
k =2mgh/ x2 = 6900 N/m
When in contact with ground the system can be considered to be an harmonic oscillator (mass on a spring) with characteristic frequency of ω = √k/m = 11rad/s
Period of oscillations is T = 2π/ω = 0.58 s
The equilibrium position (when the woman just stands on the stick is at defined by equilibrium of forces mg = kxe which give xe= mg/k = 0.085 m below the position, when the spring is fully extended.
The time when the system is in contact with ground can be well approximated as half of the period or calculated exactly as the time of the oscillation between the positions of 0.24 m above the minimum.
The time in air Ta is twice the time it takes to drop from 0.34 m to 0.155 m (0.24 – 0.085) if the approximation was used or from 0.48 m to 0.24 m for exact calculation.
Ta =2 √2h2/g= 0.39 s
Total period is Ta+ ½ T =0.68 s
One expects at least 10% uncertainty in these calculations. The graph of the position of the center of gravity as a function of time is shown below. The zero of the position scale corresponds the lowest point. The function above 0.18 m (red line) are parts of a parabola and below is a part of the cosine function.

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