JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Elasticity & SHM

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

\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).

For periodic motion of small amplitude

A

, the time period

T

of this particle is proportional to

(A) $A \frac{m}{α}$
(B) $\frac{1}{A} \frac{m}{α}$ ✓ Correct
(C) $A \frac{α}{m}$
(D) $\frac{1}{A} \frac{α}{m}$

Answer: (B)

Solution

For small amplitude oscillations near the origin with

V (x) = α x^{4}

:

The force is

F = - \frac{d V}{d x} = - 4 α x^{3}

.

For small oscillations with amplitude

A

, the restoring force at displacement

x \sim A

F \sim - 4 α A^{2} x

, giving an effective spring constant

k_{e f f} \sim α A^{2}

.

The time period

T \sim \frac{m}{k _{e f f}} \sim \frac{m}{α A ^{2}} = \frac{1}{A} \frac{m}{α}

.

Therefore, option B is correct.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Single correct option

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JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Elasticity & SHM

Solution

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