SHMHard
Question
The potential energy of a particle of mass 0.1 kg, moving along the x-axis, is given by U = 5x (x − 4) J, where x is in meters. It can be concluded that
Options
A.the particle is acted upon by a constant force
B.the speed of the particle is maximum at x = 2 m
C.the particle executes SHM
D.the period of oscillation of the particle is (π/5) sec
Solution
F = 
= − [10x − 20] = 20 − 10x
acceleration a =
= 100(2−x) = −ω2(x − 2)
a = 0 at x = 2 So V is maximum at x = 2
This is equation of S.H.M so particle executes S.H.M
also ω2 = 100 ⇒ ω = 10
T =
= π/5 sec.
= − [10x − 20] = 20 − 10xacceleration a =
= 100(2−x) = −ω2(x − 2) a = 0 at x = 2 So V is maximum at x = 2
This is equation of S.H.M so particle executes S.H.M
also ω2 = 100 ⇒ ω = 10
T =
= π/5 sec.Create a free account to view solution
View Solution FreeMore SHM Questions
In forced oscillation of a particle the amplitude is maximum for a frequency ω1 of the force, while the energy is m...The motion of a particle is represented by the function ″Asin2ωt + Bcos2ωt + Csincos″ where x, 	...The speed v of a particle moving along a straight line, when it is at a distance (x) from a fixed point of the line is g...Using a simple pendulum experiment g is determined by measuring its time period T. Which of the following plots represen...A ball is suspended by a thread of length L at the point O on the well PQ which is inclined to the vertical through an a...