The potential energy U of a body of unit mass moving in one dimensional conservative force field is given by $U = x^{2} - 4 x + 3$ . All units are in SI. For situation mark out the correct statements

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

The body will perform oscillatory motion but not simple harmonic motion

The body will perform simple harmonic motion with time period $\sqrt{2} π s$ .

If speed of the body at equilibrium position is 4m/s, then the amplitude of oscillation would be $2 \sqrt{2} m$

The body will perform simple harmonic motion about x=2 units

answer is A, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Potential energy of particle is $U = x^{2} - 4 x + 3$
$F = - \frac{d U}{d x} = \frac{- d}{d x} (x^{2} - 4 x + 3) = - (2 x - 4)$
At equilibrium F=0 at
$∴ F = - 2 x + 4 = - 2 [x + 2] = - 2 [X]$
SHM about $x = 2$ . On comparison k=2
$ω= \sqrt{\frac{k}{m}},m=unit mass$
$ω = \sqrt{2} \Rightarrow \frac{2 π}{T} = \sqrt{2} \Rightarrow T = \sqrt{2} π$
At mean position velocity is max. i.e.,
$V_{\max} = A ω = A \sqrt{2}$
$V_{\max} = 4$ then $4 = A \sqrt{2} \Rightarrow A = 2 \sqrt{2}$