# Time period of a pendulum with $3$ bobs

Here $$L$$ is the length of th whole massless rod and $$3$$ balls have been attached to the rod having lengths of $$\frac{L}{3}$$. We have to find the time period of the pendulum as a whole.

Here is my approach. Since we are used to calculating pendulum related problems for one bob,i tried to reduce the system to an equivalent one ball system.

Let us assume mass of each ball is $$m$$. Now when the pendulum is displaced with a slight angular displacement,the restoring force acting on each of the ball is $$mg\sin \theta=mg\theta$$ when $$\theta$$ is small. Since balls experience $$mg\theta$$ force in the same direction, in the equivalent one ball system,the net force on that corresponding ball will be $$3mg\theta$$. Now from hooks law, $$F=kx$$ where $$k$$ is the SHM constant and $$x$$ is displacement. Here the equivalent one ball will travserse the same angle as these $$3$$ balls, so $$\theta_{equivalent ball}$$ will be equal to $$\theta$$. Therefore $$3mg\theta=kL\theta$$ so $$k=\frac{3mg}{L}$$. And we know $$\omega^2=\frac{k}{m}=\frac{3g}{L}$$. And so $$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{L}{3g}}$$.

Is this correct?If wrong,please correct my thinking process and kindly post the actual solution.

• Are you assuming each of the sections move separately? Jan 30 at 6:12
• No if i had assumed so,i wouldn't have taken each angular displacement of the $3$ balls as $\theta$. Jan 30 at 6:51
• A double pendulum's motion is chaotic and not easy to solve. The triple pendulum is very complicated. Jan 30 at 6:59

Rods, unlike ropes, have transverse rigidity. For this reason, the transverse force acting on each ball is not equal to $$mg\theta$$. Therefore, this problem is difficult to solve based on the idea of an effective ball. To solve this problem, it is better to use the conservation of energy or equation for the angular momentum. Let's consider the later method.

For small angle $$\theta$$, torque is $$M = -mg\theta \left(\frac{L}3 + \frac{2L}3 + L \right).$$ Angular momentum is $$l = I\omega = m\left(\frac{L^2}9 + \frac{4L^2}9 + L^2\right)\dot{\theta}$$ Therefore the equation of motion in this problem has the following form: $$\dot{l} = M \quad \rightarrow \quad mL^2\frac{14}9 \ddot{\theta} = - mg2L\theta$$ From the last equation, we obtain $$\omega^2 = \frac{9g}{7L}$$

• By transverse rigidity what you mean Sir ? Jan 30 at 12:47
• I am actually not so sure about this since the actual answer isn't this.The options were $2\pi\sqrt{\frac{tL}{g}}$ where $t=1,2,3,\frac{1}{3}$. Jan 30 at 13:06
• @Orion_Pax I mean, it's impossible to bend a rod without effort, like a rope.
– Gec
Jan 30 at 15:15
• @madness In this problem, are we considering balls on a rigid rod or a flexible rope? If balls are on a rope then the system has three degrees of freedom. In this case, there are three modes of small oscillations and, possibly, three different periods. In the case of a rigid rod, you can see my solution. Maybe I'm wrong. But it is also possible that the answers are incorrect.
– Gec
Jan 30 at 15:34
• @madness There are three angles in the case of three balls on a rope. Different segments of the rope should not be parallel. Hence, there are three equations and three periods in this case. The system can be described by one angle only for balls on a rigid rod. In this case, you can see that the equations of motion of balls are not consistent without taking into account rod reaction force. A rope can not exert a corresponding reaction force on the balls to guarantee that all three segments deflect from the vertical line at the same angle.
– Gec
Jan 31 at 8:00