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
解：由题意得定解问题为[tex=17.929x7.5]fnpmC2J6JmQBLyo5NmGAzw0o8Ahogbjk6XUs7uDsJ5y/4DwobEofqKfZy5v51myyDZ+ZmVFWgBUch8xZ1/kasXj2FfUERgI48Lw8dYOzyMjJnlbOxGzvoUoSFBrTwpoPgHJtI05MTUOfzOHJLv8cAIOJmhZh0jshSUQB37pq8onMrfrFkXPgwW4O8WMYR8myOcbWwu9xLtf/pEXrtgU9bAY6h9ns/m98jAGhBQgpEa/RGAchyHu6TJzcBFiDhn5hWnsyTrA0skIGHYs71/qxg98SYpnaHW80IAFyAcvql9IIrz9d46LMKCTXXzh/csSp+CYSPYE1uz1upO4BENGYy9S0Nl23fEPhwVOYtGXFnSaxMPO66AHoyfl0EHEeiGtz[/tex]令[tex=7.429x1.357]Sn8qXtvoKx4GKBNfH/PNyejiAUxl94Bi9XhgSmu684U=[/tex]，代入泛定方程得[tex=11.0x2.429]Q03B3rKHsohITpeV7soeSpeNCOlk/q0ppnHpON/BThvWjdpL3YmOZUE/3bRwp7MGTUozfJQxpv26ES09Jj7FNKTEs76/3e5m+9IElvXPX+HC5FYmBfZAx/yuTWLkGFaB[/tex]进一步可得[tex=14.857x3.357]fnpmC2J6JmQBLyo5NmGAz/R8RKwH7hSEnm40ViGGpYeB3zq1LDOaeNlbzjlF7JoNcwkmlhrtzt7njociMNBK+M2Eky4iKXHDzLp1Ox/xo2glp9THM/sLO0dOhLg3N0HjCa0A8eYMvrgklo/Wp7k5Fk+PEtnrnA30QOAwu5QD40MZD50WmQyVsaxPISMo6x2feXC1+COs47duGzutScsPgA==[/tex]对时间变数t，解得[tex=11.0x1.357]NXlNY7vc/t6RZ3u5O41BdNRVarrbXkWh69jNAOIYMlwiCg/bLmZvARGslc+feqk0[/tex].对空间变数x，解得[tex=10.857x1.357]dH0PzSOUjmE63PcYANW+++vm+FJyo+hrN4HS/FaKYMLNTiekFil8h8NGPmi4A3nV[/tex]，则[tex=13.071x1.429]ISmHmPWsNti74QHiU00dGMvFd29QPN2BD+SZwotSiHmLQNkfDH3otUeM7RwCAqAE9UE+18vFZalWkC14LlNSyE5PPxJwREfuMiMAtAZkVww=[/tex].由X(0)=0可知C=0；由X'(l)=0可知，[tex=15.714x2.357]f7zcJ8v5YM1rDOYMvE6jwah9E869FCQwvxfWPk2HFEv04XLhcwtAlaPnTyh6b8DSxngCEci828LLvulx/XhrwQ==[/tex].[tex=31.857x6.357]c8gX0O6CKBpyqTBZ2fB4Dk8ZjaYySI/DjGHLcyqlyNnqGbTgKU7GEPpvBRFMSrjgceLsMhjyY0jK47SDhSuE80ZHk7HnUqdmu6AguD8gECpXl5sr721ozMQ6LwT9fzSXHFpT5CeCVLO79wqqbVte4Tcn1xgkCy/Q04zPoGAhMtb8sX2zPDJwRjOzpY12LK/VPAoZAH1J62Azd2T3IpKhDf9obh4hQ1IDEI79XLA3ilT+BmoSVvNaP4a2+rDpuTHugxe0EEUXgPdwjFsTnFMjSEIeIdzpt7KAikxe/pnOygGC4PZxHiZtjAmdvLcQsJcvOP2L2ICPTSf9wyykbIAoqYNmxEWZrRVWHU1Ko1Y0r6k=[/tex]由初始条件[tex=19.929x7.5]c8gX0O6CKBpyqTBZ2fB4Dkl2Dbn19MZQes4dUVJAzFnux2/grMEm8+O7IuGzxnlTAIKQehDCZpmAmcJI+kG65YRtchF04QDbMrhBgPWiHdF+I61ISfLtXhY2qbIgTJ9kMweT3dI7FNjp2McSgMNfH2dkJ16EwrjBh91pFbE8pHvRurkdQiRBH5Q0XiBAxzQ8dnjvVmD/cewEdgJMmkVYNFbz4CLpZ2Qo7WPW5CjqFw4OgNV9tf+1lmmGqaN65cLlKmemEUhNXrdlSlZnd7cqnCZGrIcym15KSekQLeE6dkhBlkl72HOWTW/kGYRjFXWQlqDZ9MV5x0tJMymG0soJTnxA94yH/yax5N5DvoZFXaTCkloEZBOnebMaJGAeKsJ80nP1tqUmk8lr/3bKxJHnlw==[/tex]则[tex=2.5x1.214]Dw3fAPVvFgPgUP5YGHMPbQ==[/tex]，[tex=34.286x13.929]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[/tex][tex=26.214x3.714]du/CgijuHviixstmM1gO8LpRElGjNXRKtX/edJSPZV/FZ9oGrmal2kzU1tZPoN5T4J2lwqmISokjjUr8wji85ZAOiOfI5smX32g+RURldfBLv0I3834mrrschzk5+csw4UEp9OKdkepEBx6VRIy4P1JBbEFUKw9T5E3+G6weitrmwHd1z+ieMsSLgkKuWF4lqicL3hUr5dANBMwXrP0cd+JLk5BVCqu+meX7imMAQW2Cr1g9KZqU5iIXI/N+Nq8r[/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。
- 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.
- 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.
- 长为 [tex=0.714x1.0]Hl8mr56J4t0Ek5ZoqbFYYg==[/tex],均匀细杆,[tex=1.857x1.0]UMz+xA7HHUC1YwcNn0Lqvw==[/tex]端固定,另一端沿杆的轴线方向拉长[tex=0.429x1.0]Q2fWySASH/4Xf2eu85OwAQ==[/tex] 静止后 (在弹性限度内 ) 突然放手，细杆做自由振动。试写出振动方程的定解条件。
内容
- 0
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.
- 1
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。
- 2
长为[tex=0.357x1.0]5vVfAZliYwqMw8JaLE+iEA==[/tex]的弦，两端固定，弦中张力为[tex=0.643x1.0]awBC2UvU2WxG45VihksPuw==[/tex]，在距一段为[tex=0.929x1.0]XQ8c0totc8uufRPOvpPxwQ==[/tex]的一点以力[tex=1.0x1.214]kLwCQn0d73TWOUUlnBVGBA==[/tex]把弦拉开，然后突然撤除这力，求解弦的振动[img=291x147]178efbe78e53a93.png[/img]
- 3
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.
- 4
一细长杆,[tex=1.857x1.0]bOlCq/PHWhsSVMaVf7Obdg==[/tex]端固定,[tex=1.714x1.0]OFSQaAQTidbnVE7HphlqPw==[/tex]端受周期力[tex=3.357x1.0]1GUhjN+YxHeRHQeqD4ARmiIX3uKXu1GpTRb4jsOBnNA=[/tex]作用.设初位移和初速度均为 0,求解此杆的纵振动问题.