"\\omega_0=\\sqrt{\\frac{k}{m}}=\\sqrt{\\frac{180}{5}}=6(rad\/s)"
(i) "A=\\frac{F_0}{m\\sqrt{(\\omega_0^2-\\omega^2)^2+4\\beta^2\\omega^2}}=\\frac{50}{5\\cdot\\sqrt{(6^2-20^2)^2+4\\cdot0.2^2\\cdot20^2}}=0.027(m)"
"\\tan\\phi=\\frac{2\\beta\\omega}{\\omega^2-\\omega_0^2}=\\frac{2\\cdot0.2\\cdot20}{20^2-6^2}=0.022\\to" "\\phi=1.26\u00b0"
(ii) "\\omega_{res}=\\sqrt{\\omega_0^2-2\\beta^2}=\\sqrt{6^2-2\\cdot0.2^2}=5.99(rad\/s)"
(iii) "A_{res}=\\frac{F_0\/m}{2\\beta\\sqrt{\\omega_0^2-\\beta^2}}=\\frac{50\/5}{2\\cdot0.2\\sqrt{6^2-0.2^2}}=4.17(m)"
Comments
Leave a comment