Potential energy of an oscillating pendulum

The following question always confuses me. for an oscillating pendulum why the potential energy is given by:

$$V = mgL(1-\cos\theta)$$

Why not $$V = mgL\cos\theta$$

Is this a convention or there something logical behind it.

I appreciate any help

• The potential energy is greater, not less, when the pendulum is higher (i.e., at greater $\theta$). So the cosine term must be negative, not positive. The additive constant is irrelevant. – G. Smith Apr 1 at 2:34

I have a simple explanation for this.Let's take the lowest point of the path of the pendulum at zero potential energy.Consider the position when bob is at angle $$\theta$$ from equilibrium position.Now the potential energy of Bob at this point has increased from zero to $$mgL(1-cos\theta)$$ • My confusion is why the bottom point $p$ is chosen as a reference? – IamNotaMathematician Apr 15 at 9:58
• We can take potential energy zero at any point.But it will be easy if we take it zero at bottom most point.If we take p=0 at top most point then,it will be negative at bottom points and if we take it zero at any other point,it will become more complicated – user227513 Apr 15 at 10:49

The gravitational potential energy lost appears as the potential energy of the pendulum (which periodically changes to kinetic energy and back), in accordance with conservation of energy. As shown in the figure, the change in potential energy = $$mgh = mgL - mgLcos \theta = mgL(1- cos \theta)$$

Whereas, $$mgLcos \theta$$ is the final gravitational potential energy with respect to ground, not change in potential energy.

I hope I clarified the confusion, but if you have any other doubt, feel free to ask in the comments

• Thank you! Just a question, I am a little bit confused because we generally talk about potential energy not the change in potential energy! – IamNotaMathematician Apr 15 at 9:56
• @IamNotaMathematician We don't measure an absolute value of potential energy, but a difference. Even $mgh$ isn't the absolute potential energy of a point. It isn't usually explicitly stated, but it is approximated for the gravitational potential energy difference between that point and ground,(or any other reference point). Even $\frac {GMm}{R}$ isn't the absolute value - it is measured considering that the potential energy at infinity is zero, so it's the difference again between the potential energy at the concerned point and infinity. – Eagle Apr 15 at 14:44

The ground is chosen as refrence for zero PE,then if the angle made by string from vertical is $$\theta$$ and length of string is $$L$$ then obviously the height of Bob from ground is $$L(1-cos\theta)$$. I cannot understand how you are reaching to $$Lcos\theta$$. Maybe the reference or sign convention is different