- 追加された行はこの色です。
- 削除された行はこの色です。
#author("2020-01-20T11:29:48+09:00","default:kouzouken","kouzouken")
#author("2020-01-20T11:30:35+09:00","default:kouzouken","kouzouken")
#contents
*平等強さの棒 [#rdc79d3a]
-http://www.loid.co.jp/html/rev/5th_day.html
${\displaystyle A_{x}=A_{0}e^{\frac{\gamma}{\sigma}x}}$が正解でいいようなので、
題意から$\sigma=\frac{P}{A_{0}}$が成り立つなら、
${\displaystyle A_{x}=A_{0}e^{\frac{A_{0}\gamma}{P}x}}$
\[ A_{x}=A_{0}e^{\frac{A_{0}\gamma}{P}x}\]
${\displaystyle U=\int_{0}^{\ell}\frac{\sigma^{2}}{2E}A_{x}dx
\[ U=\int_{0}^{\ell}\frac{\sigma^{2}}{2E}A_{x}dx
=\int_{0}^{\ell}\frac{P^{2}}{2EA_{0}^{2}} A_{0}e^{\frac{A_{0}\gamma}{P}x}dx
}$
\]