Simple Pendulum- Subjected to 2 accelerations ?

Let us analyze the Simple Pendulum Time period and mean position when subjected to both horizontal acceleration and vertical gravitational acceleration. In general we know about the simple pendulum subjected to gravitational acceleration. When it is subjected to the application of horizontal acceleration along with gravitational acceleration, the time period and mean position changes. How the values of time period and mean position change we study here.

Simple pendulum equation- horizontal and gravitational acceleration?

Here let us understand what is a simple pendulum and try to derive the simple pendulum equation with the help of simple pendulum theory. Here we consider only the influence of vertical gravitational acceleration alone. The image below details the answers considering the influence of both horizontal and gravitational accelerations.

The above image has clearly detailed all the required concepts for finding the Time period and Mean position when the system is under the influence of horizontal acceleration and gravitational acceleration. The simple pendulum formula used above has come from the simplification of the system.

Let us develop the equations and formulae used to solve the above problem step by step. There are three step calculations required here to get the final answers for Time period and mean position .

We have to find out the effective acceleration of the given system. The effective acceleration be cause of the influence of both gravitational acceleration and horizontal acceleration.

Let the accelerations be, Gravitational acceleration = g and Horizontal acceleration = a. These both act perpendicular to each other.

The effective acceleration , $g_{\mathrm{eff}}=\sqrt{g^2+a^2}$

Mean position can be obtained from the concept that at equilibrium , the resultant torque about the point of suspension will vanish. Let the angle for finding the mean position as displacement of of simple pendulum be ${\mathit{\theta }}_0$ .

By making the resultant torque about the point of suspension as zero, then

$\mathrm{mg}\sin {\mathit{\theta }}_0+\mathrm{ma}\cos {\mathit{\theta }}_0=0$

The above can be simplified as, $\sin {\mathit{\theta }}_0+\frac{a}{g}\cos {\mathit{\theta }}_0=0$

From above, we get, ${\mathit{\theta }}_0=-{\tan }^{-1}(\frac{a}{g})$ = Mean position of the equilibrium.

The displacement of the pendulum has negative sign indicating that the displacement is opposite to the horizontal acceleration direction.

Simple pendulum – Time Period

The time period T can be calculated by using the concept that for small displacements in radians from the mean position ${\mathit{\theta }}_0$ ,

$\sin \mathit{\phi }\simeq \mathit{\phi }$

Based on the above concept, and other definitions in mechanics,

$\mathit{\tau }=-{\mathrm{mg}}_{\mathrm{eff}}L\sin \mathit{\phi }={\mathrm{mL}}^2\frac{d^2\mathit{\phi }}{{\mathrm{dt}}^2}$

$\frac{d^2\mathit{\phi }}{{\mathrm{dt}}^2}+\frac{g_{\mathrm{eff}}}{L}\mathit{\phi }=0$

From above the standard SHM equation is,

$\frac{d^2\mathit{\phi }}{{\mathrm{dt}}^2}+{\mathit{\omega }}^2\mathit{\phi }=0$

hence the angular velocity $\mathit{\omega }$ is given by

$\mathit{\omega }=\sqrt{\frac{g_{\mathrm{eff}}}{L}}=\sqrt{\frac{\sqrt{g^2+a^2}}{L}}$

But $T=\frac{2\mathit{\pi }}{\mathit{\omega }}$ , hence,

$T=2\mathit{\pi }\sqrt{\frac{L}{\sqrt{g^2+a^2}}}$

Conclusion

As above for a simple pendulum, using the simple pendulum equation and simple pendulum theory,we calculated the values of Time period and Mean position when subjected to both gravitational acceleration and Horizontal acceleration are given as,

$\mathit{\omega }=\sqrt{\frac{g_{\mathrm{eff}}}{L}}=\sqrt{\frac{\sqrt{g^2+a^2}}{L}}$

and

$T=2\mathit{\pi }\sqrt{\frac{L}{\sqrt{g^2+a^2}}}$

Best of luck to everybody.

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