有一根均匀弹性细杆,长为[tex=0.357x1.0]Le5Jr6QhXJv1Yp4NjrbGVA==[/tex],一端固定,另一端受外力 [tex=4.786x1.0]06FQuz8uV5PgBt57DNOI9H7nMJczEJRN/ixtklR3kPM=[/tex] 作用.杆的初始位移与速度都为 0,求杆的纵向振动规律.
解 : 设 [tex=0.643x1.0]jDVSpgNhHe+VJmgvx3gg1Q==[/tex] 与 [tex=0.643x1.0]jLbabU9pW65GUKemsNBJWw==[/tex] 分别是细杆杨氏模量与截面积,则定解问题为[tex=13.643x5.786]7EJHVCtO2IWq3KpdB+jQsteTgYKcO485vpVNkAgPUaYwQPWJ41zrMck8sxh7lIiIUCaKHAc4BwqL7y0Id9Yy8bbSuuJg7QsdAcAS9985xXoOghxJxPujsAG8FeNHjlwx4vGLFaDbJCc8WWzldwOOCjfX2RZi6aSqQLCECcmDC0PqM/gLzz6XxcfxBYL6Gi9axgUhUZ6JEGQaYSvSMvpx7g==[/tex]对自变量 [tex=0.429x0.929]gQzDwVIykgengUJAyMAHkQ==[/tex] 取 Laplace 变换[tex=17.214x5.786]7EJHVCtO2IWq3KpdB+jQsnEtEIArvLwjH6kIe4Kz+2S/ExWA1LynGsC17AxEmStyRtG3TrRoVmgau5PPmgAx84gZ7pSrbM519y25YViBe+GFCmrU5kx1ulG0gj2jEmomeUW4ZI44aAUVGIzJE4xPhHDOn5gP5jJODA3CyUSwfTcckEPXFxEaFyllPgaOctdxLNWDAm+nU8LikeaC/podDApNndKQMvGRV+Xivu2C3DCFyvl10Clht6YUuGsd8eNeW5vV1QsnVszgIdu5MB8qD9XfmscrHl9o5xtP8jQBQ1A=[/tex]求解常微分方程可得[tex=6.929x1.357]9UVroR9HYS8USodbx+0BVse9hekLMdgDC0SC3TV0zEFI/06xJfsJ5lA+9HIRc6nLszsFxuiatqVbcDxQS9Yq4A==[/tex]代入边界条件可得出[tex=15.071x5.214]7EJHVCtO2IWq3KpdB+jQsldO2au7zElT2zIUhq2HpMbCuVs5TgOvQ3YPeul/W4L5+54FJRy3H2nZSr+ZNoQleCwLJ26cnXZwLzeij8d0ezJaolX8cmFmdtUMA687UutENf1ZSwYBq+pA7zCq/+LoeDCeVOTy9y8g5ResqZoJtmvwPlJDiv5PHjP7l54elHEKmh2MCqFcn3V717E3Yg2mEU4PccSVt+vecOXw4DaOatU=[/tex]所以 [tex=21.0x3.357]9UVroR9HYS8USodbx+0BVse9hekLMdgDC0SC3TV0zEG1OHaVcLa1VQzHukB2xO3KpB7EE5VkY0cJiFq1yo+g2M4LsLKpFiaCsMLFdHJeTslTqPZQSjEQDIeKduBA/22JuyMa46F31cc2PeN2RRF/j0xHsqIo2PXO+dx1/gUKZOVqbm55JiER34J6krrRbGFigaHPOWFIw1FfOe+lRXsBdjDQW7LzktY/XownOucdRrxa1J9LHDs4HkOwioqt+dSR+Yert/GwLeZUQIY0OwEuCA==[/tex]对上式进行逆变换就可求得杆的振动规律,此时逆变换过程复杂,又因为原定解问题是有界域问题,不妨尝试用分离变量法求解.过程如下:另解:利用分离变量法令 [tex=10.143x1.357]gJJzQyXP9MOsz/DP3ACtLoiPV2H6bO7H/GChxcSfL0o=[/tex]根据边界条件可以设 [tex=8.643x1.357]/wfcrKrfwdSdb/X5AKoJkrlsTLhqN7KBNHKFnZU9d84=[/tex]又 [tex=10.357x1.357]DdY0fZ2xFCS+f4ygLw/qr3THhGwnG6oUrjN2U15ZPdEYhv9gWHG8GPOeaJVQjkmAzFa4cPv1NmCF4qZZ1wxxgA==[/tex] 且有 [tex=8.143x2.429]Tga/ZqyGFZXNXuWkKR2c9igwW2S3LfutTLOkfOqhGGFKEjVMOSLVInHWy9tLQ5F/YBoI4WX2Vi4J1tZVHBshTg==[/tex] 可以得出[tex=9.929x2.429]z/BokFhGPYWPQ22HLs3ZWaITkF7cu30Jd9d6p8ZELTEiDfXg9TY5PgrJ4ZAxlZ1yQ0I9vBOaHCleWqSavM+JSQ==[/tex]将 [tex=8.643x1.357]/wfcrKrfwdSdb/X5AKoJkrlsTLhqN7KBNHKFnZU9d84=[/tex]代回原定解方程,并约去[tex=2.429x1.0]b/2s/+iF2Dv8QllUFLTcuw==[/tex]可得[tex=5.857x2.5]vJXo0rJ0kOT9jkxpK7SZNcpFMjrhNhYQv4/kO6N04SPH0/du73M2lJyUzMoLq0Tej5Qu5dxCfIv4HnSHalv0Lw==[/tex]上述方程的解为[tex=12.357x2.143]qr0bcm3ebXqthhiMI89y+KI2Q51mxWxLfoRjH3F5tD2IpsN38teE72lXwV0arpfXADIemo3j2FQugatDKc0q9m5l7TUWTka7RH9FcCVaKNY=[/tex], 将 [tex=3.857x1.357]z/BokFhGPYWPQ22HLs3ZWUtXB5JMMLCDcFBZbOQMtkE=[/tex] 和 [tex=5.5x2.429]rZpuY3mUDodc2P9/TRNj4PZ0oSZ795SmDBINNznAl9KHS0peIuWVPt5ms86qcHaI[/tex] 条件代入解,可以得出[tex=2.357x1.214]blWR4WPeenxo39oHlwhR1w==[/tex]和 [tex=9.714x2.929]8NZM95CgLMd2FdT/Jy50eiEM5RNiusDiLWiLrFwvw0ALf4xetEtGeve36y0f0xnnlBUFWPWlZxU/FDyYVWPqyVjEfegRSSVkBr7TSPD3I98=[/tex]所以[tex=22.786x2.929]/wfcrKrfwdSdb/X5AKoJknrah9faGYYeucDJCN64dse74TnL8GOaQoP8FpyOJivwatQ8/jIUCJxit3Tr9opteisOZBqe6rTTd2fwspKJQY9dPm2CxY5PdJYzKZCh/h2PD9pdVqVGEfs+/hUUvDOFxQJRuSwJXGK+3WLwgBMmsZJLtNvYzPtZSx/nX18ugNwa[/tex]则对应的另一个分解后的定解问题为[tex=21.5x6.357]7EJHVCtO2IWq3KpdB+jQsjkW0RGdoQTc/AXy6eRhJ8RX7xfG14TqbPdqwHD4NDfpznmtOaRIft0MBGlgPE65XuAhqP8u4FuWtPEPowGv7clg6hoHylYMsKjgTeupQ2ZXudtkkQ5qW3H0bu4Ghji5eD1aMfm/P38FYoCR5OpH/gopB2VeDTJ3QInSfBmIYld9VxSScdLJ/XS+JsmRsFigb0+jsY50KxaQjTMFfP8rLbdgB7RVZ+aYAchaNFwuixwtAmM4uTXCzl1iTDZ1LpYTKwOu3SKFZ7VEVYf4c0hEzoU=[/tex]利用分离变量法 [tex=2.786x1.0]VDuBMbUuAc4CjGM3J659+Q==[/tex],可以求出对应的特征值为[tex=13.0x2.929]mlWCpN6xvRUkipQvdD76PxMdfloJnAce3AiBcaNyboVK4qtED4/azgX7plCnRnLx5gUlb6IqwNJemZn6be+irNVg0MN3dccj6P+K3GWRnAs=[/tex]所以 [tex=10.0x2.429]j5nU4H1h1tY0E60a+8vqXgYv9DrPdqyXUFcpAI2pN6D14wh9tlNULfO7cO3kbDX2[/tex][tex=5.286x1.357]Zm2DMRan1akOb7H30GB2apNUfQb0JAi5UnZT9c4PqFiTsMYm/fs3RWlqblM608+p[/tex] 的解为[tex=19.429x2.429]jdyZVtMwa7HdidzDJZ19XFzulIcpcj+j4uzgx9VI8vOXcA558AfdV8a6NEBhF5m1iEEVB4STQC98dMUlOfdIaUh9nxfl0Uz52iRIUUlVfPt/vf4LiBK+LxlEOV8Fu9ZR[/tex]则[tex=29.5x3.286]XEEW94QdxdMhi+p47kY8uJHO0RIVnhy1dQqRBLVZWHZvEtGYHUJhhDRzARQogSjj7Zx6ePu/eDR/CRj5vzREWhysMuIWZftXsIHpt6KUw2AXyfCvYQRV8U8PQo3dBKpqJe9pIdUMQlBEn6hHFCUMPLbW2AkmQ3SFLJYORSgPdXOQOqlQ+fybhjd43ryBQNWkl5VdXQyBvYPWOf07OO60KQ==[/tex]根据初始条件 [tex=4.214x1.357]4oUxs++SlwZ3nv53PfliZQ==[/tex],可得出 [tex=2.429x1.214]BqxhHxMl2Nwyh8H2HNlNSw==[/tex][tex=29.214x3.286]m46ErxkVsqfNGogpX9OK4Iwq4gRRMCDTvCJrkQWgEfQ13WR9zAHf/ogzSEwfWsPcO8KrylHLsLMxIqnkE6LhPAWZ4IZoeTHaHvEonL+wmN9TGzId6XXPZW89P79mXAjZ34z4Vvz2c+dkKnEK4yu5ivvSyWFoNCXlo2x8OhPHO9cFtB6VAKheFtkRv8S5SQv7ZTbW7kPhirGUJgx3O9mOmqMUtdlW3QqeHMO7pPBIJdKVY53CyV3VZdNYeJUe2MvA[/tex]根据傅里叶变换可得[tex=26.643x6.357]qeiYnKXLEhyhuGRg8yLtr19JB1QHzJQAYZwVMcmo5CNwYAhX9G3S/SBvPFOWit/IVV5T30jcyt2I/DunvpQ14YQEmCxW5B0v1FCLHVBzDH8a7CMgNUeHDAvJQx+QNev3mpJPbvRD8vmfW/zQypNrikeo0xZVMnTTEDE+gu9dcDbzko9puROu20xz1WDbtIhIp9x3DdV14c1ZxrdxCUiB5k7bRsXRt/u6Wi8XDDuFfY4vBBPXU2GlF9bH29wzYaoAqTu0mDpCdHQGIsrgWqbeHkJLXVSFhbVF1rLr03rNDfXZHoFIXZXcoLheGGQjc0Xm362CWTYEjBh6aLoBtthtO/4FbXkmZ7Cht3vAZg2nbqseHX8ROwL8dV8xsDTGAzqCxgWum0EiP0JyRRDpSo7RUVFwRLLrDW5dAwxX/X4EiG+ySHNJZVSAzPW93ed/bvyfFpMq+8VlTekXfVaUvAGeWX7iO1tWJdTPtcVjwjT/4Fw=[/tex]对于积分[tex=25.571x3.071]Q+YOETNz6AgYUXGIP1wIVM8WIGG47IJegOTuSRgDcrClIekMgiv2jKs3LpTdBeq7oorHT9qLbqot2IPdIExnG6F32Oj9MjGycIR3lkEQFu9M5xEJwsvO5HK9itYMl2Fti8eX+SGK2r90uVsflw5ph26B3WzbaKaVjbRTGvDCyiPIrhYwP361/vI/5ETPAPAR2m2hkbK9p9yfvUPjf90JAMuSeNj09rzS0J0mM18jLgRxQriZ6AstWr5GuFR/oNV6FkKJO78SRcoXrQ/GV1lEb6peENR/7xGVB5PwPtSqEkE=[/tex]则[tex=26.714x12.5]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[/tex]所以[tex=36.357x10.357]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[/tex]故原定解问题的解为[tex=39.643x3.286]zPmEqyAw4LoF5lzyuTlUswBbt1HJTJpiHaD6KY6yv3d+4McSOLOIU8EgvBCEIWaZmCXLui8iGbVJWXMexBZ5yGgCIHuEBTOSTArtlGf7a2KsJ08QGphWoAskHjvfUDU7j+v4+KdhCwc0Lx9nAUHxtMbQraoAleblRg8mqcrkrOdcsmtF5koKlPgciuIDW/RXyr8ChcPNsN1NL/jKqpwyIQdogkkIu4vesUl55I/nfO3/XjbyTn8H2InVNU724xIxol9XqbYdEVN1SP3AGl4Tk5njdFXF1zGYfJlJulBSsC7NuEQbmt8HqB0uN9Az4kL/bjacQmTFWNR53VYIoHOXKpI8HUPAE8V8xlvi2DG6cOOIiBz6h70H7zPdElSC29ozlgrZgtJ5gqa7GcOjFLtbeHt6nFkSa5TA/O8GpthNbps=[/tex]
举一反三
- 长为[tex=0.357x1.0]5vVfAZliYwqMw8JaLE+iEA==[/tex]的均匀杆，一端固定,另一端在纵向力[tex=6.143x1.357]zZbR7KX3YYCMAfDkhy05VeIWnS8yBhyzt0vYIWTeRdg=[/tex]长期作用下.求解杆的恒定纵振动.
- 一均匀杆原长是[tex=0.357x1.0]Le5Jr6QhXJv1Yp4NjrbGVA==[/tex]，一端固定，另一端沿杆的轴线方向被拉长[tex=0.5x0.786]rCTQ93hYjIOF3vc8FasIqg==[/tex]而静止，突然放手任其振动，试建立振动方程和定解条件.
- 一均匀杆的原长为 [tex=0.357x1.0]5vVfAZliYwqMw8JaLE+iEA==[/tex] , 一端固定，另一端沿杆的轴线方向拉长 [tex=0.5x0.786]rCTQ93hYjIOF3vc8FasIqg==[/tex] 而静止，突然放手任其 振动，试建立振动方程与定解条件。
- 研究长为[tex=0.357x1.0]Le5Jr6QhXJv1Yp4NjrbGVA==[/tex]，一端固定，另一端自由，初始位移为[tex=1.214x1.0]oRQs3fUc5jUXOKKnlCZAtw==[/tex]而初始速度为零的弦的自由振动情况。
- 设有长为[tex=0.357x1.0]Le5Jr6QhXJv1Yp4NjrbGVA==[/tex]的均匀细杆,一端保持温度为[tex=0.929x1.0]M6rCjWOyyOXOB1PmbinM2A==[/tex],另一端绝热.杆的初温为 0 .求杆中温度的分布和变化.
内容
- 0
长为[tex=0.357x1.0]5vVfAZliYwqMw8JaLE+iEA==[/tex]的杆,一端固定,另一端受力[tex=1.0x1.214]kLwCQn0d73TWOUUlnBVGBA==[/tex]而伸长.求解杆在放手后的振动
- 1
一长为[tex=0.357x1.0]Le5Jr6QhXJv1Yp4NjrbGVA==[/tex] 、横截面积为[tex=0.643x1.0]jLbabU9pW65GUKemsNBJWw==[/tex]的均匀弹性细杆，已知[tex=1.857x1.0]bOlCq/PHWhsSVMaVf7Obdg==[/tex]的一端固定，在[tex=1.714x1.0]OFSQaAQTidbnVE7HphlqPw==[/tex]的一端在杆轴方向上受拉力[tex=0.857x1.0]WBOxEEx6dPfNM3eGriw9WQ==[/tex]作用而平衡，在[tex=1.643x1.0]MVeOYouc7e3FvU1m5bCV6w==[/tex]时撤去外力，并忽略重力的作用，试列出杆的纵振动所满足的方程、边界条件和初始条件.
- 2
设初始温度为零，长为[tex=0.357x1.0]Le5Jr6QhXJv1Yp4NjrbGVA==[/tex]的均匀细杆，当杆的一端温度为[tex=0.929x1.0]M6rCjWOyyOXOB1PmbinM2A==[/tex]，而另一端及杆的侧面对于周围介质热绝缘时，求杆中的温度分布.
- 3
一均匀带电细杆,长为[tex=4.5x1.214]rFNrkr54zZGMn6USFh7sjA==[/tex] 线电荷密度 [tex=7.929x1.5]ZZDwZMDukPt3ZUs1YYMMLojQ1WM31sgLspce/kOtfmn1SD/2co8gCAtXYK3Z8cNe[/tex], 求:(1) 细杆延长线上与杆的一端相距[tex=3.929x1.0]+WJka6cvYgcAP7RQxEzcEg==[/tex]处的电势;(2) 细杆中垂线上与细杆相距 [tex=3.857x1.0]fy3zkdxZvy5DJcdOJaUoqA==[/tex]处的电势。
- 4
长为[tex=0.357x1.0]Le5Jr6QhXJv1Yp4NjrbGVA==[/tex]的杆，侧面和[tex=1.857x1.0]bOlCq/PHWhsSVMaVf7Obdg==[/tex]端绝热，另一端[tex=1.714x1.0]OFSQaAQTidbnVE7HphlqPw==[/tex]与外界按Newton冷却定律交换热量（设外界温度为0），初始时刻杆内温度为常数[tex=0.929x1.0]M6rCjWOyyOXOB1PmbinM2A==[/tex]，求杆内温度分布.