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JEE Advance - Physics (2010 - Paper 1 Offline - No. 22)

When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, 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 kx2 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) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).

When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, 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 kx2 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) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).

When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, 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 kx2 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) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).

For periodic motion of small amplitude A, the time period T of this particle is proportional to
$$A\sqrt {m/\alpha } $$
$${1 \over A}\sqrt {m/\alpha } $$
$$A\sqrt {\alpha /m} $$
$${1 \over A}\sqrt {\alpha /m} $$

Explanation

As $$V = a{x^4}$$

$$[\alpha ] = {{[V]} \over {[{x^4}]}} = {{[M{L^2}{T^{ - 2}}]} \over {[{L^4}]}} = [M{L^{ - 2}}{T^{ - 2}}]$$

By method of dimensions,

$$\left[ {{1 \over A}\sqrt {{m \over \alpha }} } \right] = {{{{[M]}^{1/2}}} \over {[L]{{[M{L^{ - 2}}{T^{ - 2}}]}^{1/2}}}} = [T]$$

Only option (b) has the dimensions of time.

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