IIT-JEE 2010 · Paper 1 · Physics Q71

← IIT-JEE 2010 · Paper 1

PhysicsSingle correct+3 / -1Question 71 of 84

Passage

Paragraph for Questions 70 to 72

When a particle of mass m moves on the x-axis in a potential of the form $V(x)=kx^2$ , it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\dfrac mk}$ , as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of $x=0$ in a way different from $kx^2$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is $V(x)=\alpha x^4$ ( $\alpha>0$ ) for $|x|$ near the origin and becomes a constant equal to $V_0$ for $|x|\ge X_0$ (see figure).

V(x) against x: zero at the origin, rising smoothly on both sides (like αx⁴) and levelling off at the constant V₀ for |x| ≥ X₀.

For periodic motion of small amplitude A, the time period T of this particle is proportional to

A\sqrt{\dfrac m\alpha}

\dfrac1A\sqrt{\dfrac m\alpha}

A\sqrt{\dfrac\alpha m}

\dfrac1A\sqrt{\dfrac\alpha m}

Answer and worked solution

Official answer: B

Dimensional analysis. For $V=\alpha x^4$ , the energy at amplitude $A$ is $E=\alpha A^4$ , so the period can depend only on $m$ , $\alpha$ and $A$ . With $[\alpha]=\mathrm{ML^{-2}T^{-2}}$ :
$T\propto m^a\alpha^bA^c:\quad \mathrm M:\ a+b=0,\quad \mathrm L:\ -2b+c=0,\quad \mathrm T:\ -2b=1.$
So $b=-\frac12$ , $a=\frac12$ , $c=-1$ : $T\propto\dfrac1A\sqrt{\dfrac m\alpha}$ .

(Larger amplitude means a steeper restoring force, so a shorter period.)

(B)

Topics:Dimensional analysis Simple harmonic motion

← Previous question Next question →

IIT-JEE 2010 · Paper 1, Physics question 71