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
108 views
in Physics by (76.6k points)
closed by
A particle moving on x - axis has potential energy `U = 2 - 20x + 5x^(2)` joule along x - axis. The particle is relesed at `x = -3`. The maximum value of `x` will be (`x` is in metre)
A. 5m
B. 3m
C. 7m
D. 8m

1 Answer

0 votes
by (76.3k points)
selected by
 
Best answer
Correct Answer - C
`U=2-20x+5x^(2)`
`F=-(dU)/(dx)=20-10 x `
At equilibrium position , F=0
`20-10 x=0`
`rArr x=2`
Since particle is released at x=-3, therefore amplitude of particle is 5.
image
It will oscillate about x=2 with an amplitude of 5.
`:. ` maximum value of x will be 7.

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

...