Question 365: Oscillations

Question

The period of oscillation of a simple pendulum of length $L$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\alpha$, is given by

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

$2\pi \sqrt {{L \over {g\cos \alpha }}} $

Answer

$2\pi \sqrt {{L \over {g\sin \alpha }}} $

Answer

$2\pi \sqrt {{L \over g}} $

Answer

$2\pi \sqrt {{L \over {g	an \alpha }}} $

Oscillations

The period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination \alpha, is given by

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