If the total energy of the particle is E, it will perform periodic motion only if [2010]

Question

When a particle of mass $m$ moves on the $x$ -axis in a potential of the form $V (x) = k x^{2}$ it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{m}{k}},$ 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 $k x^{2}$ 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 $V (x) = α x^{4} (α > 0)$ for $| x |$ near the origin and becomes a constant equal to $V_{0}$ for $| x |$ $\geq X_{0}$ (see figure). [2010]

Q. If the total energy of the particle is E, it will perform periodic motion only if

Smart Achievers · Accepted Answer

(3)

The particle will not perform oscillations if $x$ Therefore, $E > 0 .$ If $E = V_{0},$ the potential energy will become constant as depicted in the graph given. In this case also the particle will not oscillate.

$∴$ $E > V_{0}$

$∴$ $V_{0} > E > 0$

Solution

1 $E < 0$

2 $E > 0$

3 $V_{0} > E > 0$

4 $E > V_{0}$

Ans.
(3)

The particle will not perform oscillations if $E \leq 0 .$ Therefore, $E > 0 .$ If $E = V_{0},$ the potential energy will become constant as depicted in the graph given. In this case also the particle will not oscillate.

$∴$ $E > V_{0}$

$∴$ $V_{0} > E > 0$

Want Us to Email you About Special Offers & Updates?

Solution

1 E<0

2 E>0

3 V0>E>0

4 E>V0