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
83 views
in Physics by (91.2k points)
closed by
A particle is `SHM` is discribed by the displacement function `x(t) = a cos (Delta omega + theta)` If the initial `(t = 0)` position of the particle `1cm` and its initial velocity is `picm//s` The angular frequency of the particle is `pi rad//s`, then its amplitude is
A. `1cm`
B. `sqrt(2)cm`
C. `2 cm`
D. `2.5 cm`

1 Answer

0 votes
by (91.8k points)
selected by
 
Best answer
Correct Answer - B
`x = a cos(omega t +theta)`…(i)
And `v = (dx)/(dt) = - aomegasin(omega t + theta)`…(ii)
Given at `t = 0 , x = 1 cm and v = pi` and `omega = pi`
Putting these value in equation (i) and (ii) , are will get
`sin theta = (-1)/(a)` and `cos theta = (1)/(A)`
`rArr sin^(2) theta + cos^(2) theta = (-(1)/(a))^(2) +((1)/(a))^(2) rArr a = sqrt(2) cm`

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

...