Question

急求这道物理题答案！要讲解！！！

当温度为20.0摄氏度时，空气柱产生共振。频率
生产的是用计算机记录的；但是，有一个问题，我们无法判断
记录了哪些谐波。我们记录了连续泛音860赫兹，1204赫兹
和1548赫兹。回答下列问题：
a、 基频是多少？
b、 管子是开着的还是关着的？
c、 管子的长度是多少？
英文原题：
An air column is made to resonate when the temperature is 20.0o C. The frequencies
produced are recorded using a computer; however, there is a problem and we cannot tell
which harmonics are being recorded. We record successive overtones of 860 Hz, 1204
Hz, and 1548 Hz. Answer the following questions:
a. What is the fundamental frequency?
b. Is the tube open or closed?
c. What is the length of the tube?

Answer 1

向东的位移是S1＝V1*t1＝5*6＝30米
向北的位移是S2＝V2*t2＝3*4＝12米
这10秒内发生的位移是　S＝根号（S1^2＋S2^2）＝根号（30^2＋12^2）＝32.31米
所求的平均速度大小是　V度＝S / t＝S /(t1＋t2)＝32.31 / 10＝3.231 m/s
（平均速度方向是东偏北）
所求的平均速率是　V率＝（S1＋S2）/ t＝（S1＋S2）/(t1＋t2)＝（30＋12）/ 10＝4.2 m/s