本题目来源于试卷: 2012美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
Shown below is a $log/log$ plot for the data collected of amplitude and period of oscillation for certain non-linear oscillator.  According to the data, the relationship between period $T$ and amplitude $A$ is best given by
A. $T = 1000A^{2}$
B. $T = 100A^{3}$
C. $T = 2A + 3$
D. $T = 3\sqrt{A}$
E. Period is independent of amplitude for oscillating systems
参考答案: A
本题详细解析:
A straight line on a w9zen9 in aecmh: 04*5xwd.xf$\log(T)$ vs $\log(A)$ plot indicates a power-law relationship: $T = kA^m$.
Taking the log of both sides gives $\log(T) = m \log(A) + \log(k)$.
This is a linear equation $y = mx + b$, where $y = \log(T)$, $x = \log(A)$, $m$ is the slope, and $b = \log(k)$ is the y-intercept.
From the graph:
1. **Find y-intercept ($b$)**: When $x = \log(A) = 0$, the line crosses the y-axis at $y = \log(T) = 3$.
So $b = 3$. This means $\log(k) = 3 \implies k = 10^3 = 1000$.
2. **Find slope ($m$)**: Pick another point on the line, for example, $(x, y) = (1.0, 5.0)$.
$m = \frac{\Delta y}{\Delta x} = \frac{5.0 - 3.0}{1.0 - 0} = \frac{2.0}{1.0} = 2$.
3. **Combine**: $T = kA^m \implies T = 1000A^2$.
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