Question

物理题目。求解。

jose，whose mass is 90 KG, has just completed his first bungee jump and is now bauncing up and down at the end of the cord。 His osollations have an initial amplitude of 13M and a period of 4.0s。hint：Although not entirely realistic，treat the bungee cord as an ideal spring that can be compressed to a shorter length as well as stretched to a longer length。
a. what is the spring constant of the bungee cord?
b. what is Jose's maximum speed while oscillating ?
c.Form what height above the lowest point did Jose jump?
d. if the damping constant due to air resistance is 6.0 kg/s, how many oscillations will Jose make before his amplitede ihas decreased to 5.0 m?

Answer 1

琼斯，他的质量是90KG，他刚刚做了蹦极跳，现在正在绳末尾上下的蹦达。他振动的起始振幅是13M，周期为4秒。提示：尽管不完全符合实际，把蹦极的绳子视为理想弹性的。
a。蹦极绳子的弹性系数是多少？
b。琼斯在振动时候的最大速度是多少？
c。计算琼斯起跳点与最低点的高度差。
d。如果空气阻力的阻尼系数是6kg/s，在他的振幅降低至5m之前，琼斯做了多少上次振动？
解:
a. T=2 π (m/k)^1/2 其中T为周期，k为弹性系数

b. 振动方程可以表示为:y(t)=Asinωt
速度 v(t)=dx/dt=Aωcoswt
Vmax=Aω,其中A为振幅13m，ω为2π/T

c. mgΔh1 =1/2 mv^2
其中 v用Vmax代,求出Δh1

d. mgΔh2=xFf
其中Δh2=13m-5m=8m,Ff为空气阻力
带入上式可求出x，即琼斯在振幅降至5m时他在空中运动过的距离。
后面不会做了，惭愧。。。

Answer 2

这貌似比较难的是英文，看不懂。。。如果你把它翻译一下，我就能帮你解答一下。。。