The displacement (x) of a particle as a function of time (t) is given by x = a sin (bt + c)where a, b and c are constants of motion. Choose the correct statements from the following.
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
The motion repeats itself in a time interval of 2π/b
b
The energy of the particle remains constant.
c
The velocity of the particle is zero at x=±a
d
The acceleration of the particle is zero at x=±a
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
The motion of the particle is simple harmonic. The displacement at time t is: x = a sin (bt + c)Therefore, displacement at time (t+(2π/b)) isx at t+2πb=asinbt+2πb+c=asin[bt+c+2π]=asin(bt+c)=x( at time t)Hence, statement (a) is correct.Statement (b) is also correct since the same displacement is recovered after a time interval of (2π/b). Statement (c) is correct because the velocity is zero when the displacement = ±the amplitude, i.e.. at the extreme ends of the motion. Statement (d) is incorrect, because the acceleration is maximum (in magnitude) at r =±a.