平成 28 年 7 月 11 日 の解答 課題 1 𝑋0 = |𝐺(𝑗𝜔)| × 𝐹0 = 1 √𝐷 2 𝜔 2 𝜑 = ∠𝐺(𝑗𝜔) = − tan−1 + 𝐾2 × 𝐹0 𝐷𝜔 𝐾 ただし K = 100 [N/m], D = 1 [Ns/m] Fo = 10 [N] 周波数 10 [Hz] 周波数が 10Hz より ω = 2𝜋 × 10 [ 𝑋0 = 1 √12 × (2𝜋 × 10)2 + 1002 𝜑 = − tan−1 𝑟𝑎𝑑 𝑠 ] × 10 = 0.0847 [m] 1 × (2𝜋 × 10) = −0.561 [rad] (−32.1[°]) 100 周波数が 0 では ω = 2𝜋 × 0 = 0 𝑋0 = 1 √12 × 02 + 1002 𝜑 = − tan−1 × 10 = 0.1 [m] 1 × (2𝜋 × 0) = 0 [rad] ( 0 [°]) 100 課題 2 𝐹0 𝐷 1 1 𝑎 𝑏 (𝑠+𝐾/𝐷 × 𝑠 ) = 𝑠 + 𝑠+𝐾/𝐷 𝑎 = lim 𝑠→0 𝑏= と置くと 𝐹0 1 1 𝐹0 1 𝐹0 ( × ) × 𝑠 = lim × = 𝑠→0 𝐷 𝑠 + 𝐾/𝐷 𝑠 𝐷 𝐾/𝐷 𝐾 lim 𝑠→−𝐾/𝐷 𝐹0 1 1 𝐹0 1 𝐹0 ( × ) × (𝑠 + 𝐾/𝐷) = × =− 𝐷 𝑠 + 𝐾/𝐷 𝑠 𝐷 −𝐾/𝐷 𝐾 従って 𝐹0 1 1 𝐹0 1 1 ( × )= ( − ) 𝐷 𝑠 + 𝐾/𝐷 𝑠 𝐾 𝑠 𝑠 + 𝐾/𝐷
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