Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
79 views
in Physics by (47.2k points)
closed by
सरल आवर्त गति में किसी कण का वेग ` v = 10 cos (2t +pi/4)` मीटर /सेकण्ड है । कण की गति का आयाम , आवर्तकाल , आवृत्ति तथा प्रारम्भिक कला ज्ञात कीजिये ।

1 Answer

0 votes
by (67.5k points)
selected by
 
Best answer
कण का समय - वेग समीकरण
` upsilon = 10 cos (2y +pi/4) " "` …(1)
सरल आवर्त गति में समय - वेग समीकरण
` upsilon = A omega cos (omega t + phi)" " ` …(2)
समीकरण (1 ) व (2) की तुलना करने पर ,
अधिकतम वेग ` upsilon_("max") = A omega = 10` मीटर /सेकण्ड
कोणीय आवृति `omega = 2 ` रेडियन/सेकण्ड
अतः कण की गति का आयाम `A = (v_("max"))/omega = 10/2 =5` मीटर
आवर्तकाल `T = (2pi)/omega = (2pi)/2 = pi` सेकण्ड
आवृति` f = 1/T = 1/pi` हर्टज
समीकरण ( 1) से ,
कण की कला ` = 2t +pi/4`
` :. " प्रारंभिक कला " = pi/4`

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...