本题目来源于试卷: 2007美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
A thin, uniform rod + qbvtvz814f :0 lpqqcd2zkg/4jq fq f 7oq4t:3xb*qyhas mass $m$ and length $L$. Let the acceleration due to gravity be $g$. Let the rotational inertia of the rod about its center be $md^{2}$. The rod is suspended from a point a distance $kd$ from the center, and undergoes small oscillations with an angular frequency $\beta\sqrt{\frac{g}{d}}$. Find the maximum value of $\beta$.
A. $1$
B. $\sqrt{2}$
C. $1/\sqrt{2}$
D. $\beta$ does not attain a maximum value
E. none of the above
参考答案: C
本题详细解析:
From problem 32, we havn ) d)cmjivr.h+; k*she $\beta = \sqrt{\frac{k}{1+k^2}}$.
To find the maximum value of $\beta$, we can find the maximum of $\beta^2 = \frac{k}{1+k^2}$.
We take the derivative of $\beta^2$ with respect to $k$ and set it to zero.
$\frac{d(\beta^2)}{dk} = \frac{d}{dk} \left( \frac{k}{1+k^2} \right)$
Using the quotient rule:
$\frac{d(\beta^2)}{dk} = \frac{(1)(1+k^2) - (k)(2k)}{(1+k^2)^2} = \frac{1+k^2-2k^2}{(1+k^2)^2} = \frac{1-k^2}{(1+k^2)^2}$.
Set the derivative to zero: $\frac{1-k^2}{(1+k^2)^2} = 0$.
This requires the numerator to be zero: $1-k^2 = 0 \implies k^2 = 1 \implies k=1$ (since $k$ represents a distance).
Now, substitute $k=1$ back into the expression for $\beta$ to find the maximum value:
$\beta_{max} = \sqrt{\frac{1}{1+1^2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
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