ÌâÄ¿ÄÚÈÝ

2£®ÈçͼÊÇÑо¿ÎïÌå×öÔȱäËÙÖ±ÏßÔ˶¯µÄʵÑéµÃµ½µÄÒ»ÌõÖ½´ø£¨ÊµÑéÖдòµã¼ÆÊ±Æ÷Ëù½ÓµÍѹ½»Á÷µçÔ´µÄƵÂÊΪ50Hz£©£¬°´ÕÕ´òµãµÄÏȺó˳ÐòÒÀ´Î¼ÇΪ0¡¢1¡¢2¡¢3¡¢4¡¢5¡¢6£¬ÏàÁÚÁ½¼ÆÊýµã¼ä»¹ÓÐ4¸öµãûÓл­³ö£¬²âµÃx1=5.18cm£¬x2=4.40cm£¬x3=3.62cm£¬x4=2.78cm£¬x5=2.00cm£¬x6=1.22cm£®

£¨1£©´òµã¼ÆÊ±Æ÷µÄ¹¤×÷µçÁ÷Êǽ»Á÷£¨Ñ¡Ìî¡°½»Á÷¡±»ò¡°Ö±Á÷¡±£©£»
£¨2£©ÏàÁÚÁ½¼ÆÊýµã¼äµÄʱ¼ä¼ä¸ôΪ0.1s£®
£¨3£©ÎïÌåµÄ¼ÓËÙ¶È´óСa=0.78m/s2£¬£¨½á¹û±£Áô2λСÊý£©·½ÏòA¡úB£¨Ñ¡Ìî¡°A¡úB¡±»ò¡°B¡úA¡±£©£®
£¨4£©´òµã¼ÆÊ±Æ÷´ò¼ÆÊýµã3ʱ£¬ÎïÌåµÄËÙ¶È´óСv3=0.32m/s£®£¨½á¹û±£Áô2λСÊý£©

·ÖÎö ¼ÆÊ±Æ÷ʹÓý»Á÷µçÔ´£¬¶ø½»Á÷µçÔ´µÄƵÂÊΪ50Hz£¬½áºÏÏàÁÚÁ½¼ÆÊýµã¼ä»¹ÓÐ4¸öµãûÓл­³ö£¬¼´¿ÉÇó½âʱ¼ä¼ä¸ô£»
¸ù¾ÝÔȱäËÙÖ±ÏßÔ˶¯µÄÍÆÂÛ¹«Ê½¡÷x=aT2¿ÉÒÔÇó³ö¼ÓËٶȵĴóС£¬
¸ù¾ÝÔȱäËÙÖ±ÏßÔ˶¯ÖÐʱ¼äÖеãµÄËٶȵÈÓڸùý³ÌÖÐµÄÆ½¾ùËÙ¶È£¬¿ÉÒÔÇó³öС³µ¾­¹ý¼ÆÊýµã3ʱµÄ˲ʱËÙ¶È´óС£®

½â´ð ½â£º£¨1£©´òµã¼ÆÊ±Æ÷µÄ¹¤×÷µçÁ÷Êǽ»Á÷µç£»
£¨2£©´òµã¼ÆÊ±Æ÷µÄ´òµãÖÜÆÚΪ0.02s£¬ËùÒÔÏàÁÚÁ½¼ÇÊýµã¼äµÄʱ¼ä¼ä¸ôΪ£ºT=5¡Á0.02=0.1s£®
£¨3£©¸ù¾ÝÖð²î·¨ÓУºa=$\frac{{s}_{6}+{s}_{5}+{s}_{4}-£¨{s}_{3}+{s}_{2}+{s}_{1}£©}{9{T}^{2}}$=$\frac{£¨1.22+2.00+2.78£©-£¨3.62+4.40+5.18£©}{9¡Á0£®{1}^{2}}$¡Á10-2=-0.78m/s2£¬
¸ººÅ±íʾÓëÔ˶¯·½ÏòÏà·´£¬¼´¼ÓËÙ¶È·½ÏòΪA¡úB£®
£¨4£©¸ù¾ÝÔȱäËÙÖ±ÏßÔ˶¯ÖУ¬Öмäʱ¿ÌµÄËٶȵÈÓڸùý³ÌÖÐµÄÆ½¾ùËÙ¶ÈÓУºv3=$\frac{{s}_{3}+{s}_{4}}{2T}$=$\frac{3.62+2.78}{2¡Á0.1}¡Á1{0}^{-2}$=0.32m/s£®
¹Ê´ð°¸Îª£º£¨1£©½»Á÷£¬£¨2£©0.1£»£¨3£©0.78£»A¡úB£»£¨4£©0.32£®

µãÆÀ ±¾Ìâ½èÖúʵÑ鿼²éÁËÔȱäËÙÖ±ÏߵĹæÂÉÒÔ¼°ÍÆÂÛµÄÓ¦Óã¬ÔÚÆ½Ê±Á·Ï°ÖÐÒª¼ÓÇ¿»ù´¡ÖªÊ¶µÄÀí½âÓëÓ¦Óã¬Ìá¸ß½â¾öÎÊÌâÄÜÁ¦£¬×¢ÒâÓÐЧÊý×Ö±£Áô£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø