Question 364: Oscillations

Question

A particle executes simple harmonic motion between x = - A to x = + A. The time taken for it to go from 0 to $${A \over 2}$$ is T
1
and to go from $${A \over 2}$$ to A is T
2
. Then

Accepted Answer

The potential energy $U(x)=k\left[1-e^{-x^2}ight]$, varies with $x$ as shown in the figure.
The potential energy has a local minimum $\left(\frac{\mathrm{d} U}{\mathrm{~d} x}=ight.$ $0, \frac{\mathrm{~d}^2 U}{\mathrm{~d} x^2}>0$ ) at stable equilibrium and local maximum $\left(\frac{\mathrm{d} U}{\mathrm{~d} x}=0, \frac{\mathrm{~d}^2 U}{\mathrm{~d} x^2}<0ight)$ at unstable equilibrium. It is given that potential energy has a stable equilibrium at $x=0$. When particle is displaced from the origin, the force $F=-\mathrm{d} U / \mathrm{d} x$ brings it back. This force points towards the origin. There is no unstable equilibrium.
Total mechanical energy, $U+K=k / 2$, gives kinetic energy of the particle as $K=k / 2-U$, which is maximum at $x=0$. The force on a particle of mass $m$ placed near $x=0$ is given by
$$ \begin{aligned} F & =-\frac{\mathrm{d} U}{\mathrm{~d} x}=-2 k x e^{-x^2}=-2 k x\left(1-\frac{x^2}{2!}+\cdotsight) \\ & \approx-2 k x \end{aligned} $$
The force $F$ is proportional to $x$ and points towards $x=$ 0 . Thus, particle executes SHM with angular frequency $\omega=\sqrt{2 k / m}$.

Answer

T
1
< T
2

Answer

T
1
> T
2

Answer

T
1
= T
2

Answer

T
1
= 2T
2

Oscillations

A particle executes simple harmonic motion between x = - A to x = + A. The time taken for it to go from 0 to ${A \over 2} is T 1 and to go from {A \over 2}$ to A is T 2 . Then

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