0
$\begingroup$

The time period of a pendulum is given by $$T=2\pi\sqrt{\frac{l}{g}}$$

Will the time period change if a constant horizontal force acts on the pendulum? For example, if a force $F$ acts on the Bob along the horizontal. Or applying an electric field along the horizontal and giving the Bob a charge.

I think the time period won't change but the equilibrium position will change. The only restoring force is gravity. I tried to prove it:

For small displacement of $\theta$ about point of suspension, the torque about the point of suspension is $$\tau=F\cos\theta\cdot l - mg\sin\theta\cdot l$$ $$I\alpha=F\cos\theta\cdot l - mg\sin\theta\cdot l$$ For small angular displacement, $\sin\theta=\theta$ $$I\alpha=F\cos\theta\cdot l - mg\theta\cdot l$$ For SHM, $\alpha$ should be proportional to $\theta$. The only term that obeys the SHM rule is $mgl\theta$.

$\endgroup$
2
$\begingroup$

Gravity exerts a constant force downwards. If you apply a second constant force sideways you are in effect rotating the gravitational force:

Pendulum

where the angle $\theta$ is given by:

$$ \tan\theta = \frac{F_\text{ext}}{F_g} $$

The modulus of the net force is given by the usual Pythagoras rule:

$$ F_\text{net}^2 = F_g^2 + F_\text{ext}^2 $$

So if you imagine mentally rotating the whole experiment an angle $\theta$ anticlockwise you would have the pendulum handing vertically downwards in an effective gravitational field of $F_\text{net}$. This should make it obvious how the period changes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.