The Pendulum & SHM #7

The pendulum is famous example of an oscillating body doing SHM or simple harmonic motion. Well, this is true only if the angle of displacement of the pendulum is small. Are you "PHYSICS READY?" : https://the-science-cube.teachable.com/

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So that brings us to the famous example of an oscillating pendulum that does simple harmonic motion.
Well, we will see that this is true only if the angle of displacement of the pendulum is small.
So this is our pendulum that extends an angle theta with the vertical. The forces acting on the ball of the pendulum are one, the force of gravity acting in downward direction and the tension in the string that tends to pull the ball in upward direction. And lets say the length of the string is “l”
The ball will arc and oscillate. Let us label this as motion in X direction and this in Y direction.
If you decompose tension T in its respective components, you have T Cos theta as the vertical component and T Sin theta as the horizontal component.
Now we know that Newtons 2nd law says that F= ma
So lets go ahead and write the force equations for X and Y directions
In X direction the only force we can see is T Sin theta so we can write
Ma = -T sin theta = - Tx/l
In Y direction we can write
Ma = T cos theta – mg
And now we make two assumptions
1. Small angle approximation – that is the displacement of the pendulum will be through a small angle only since for larger angles, the pendulum will not exhbit SHM. Infact, smaller the angle, more perfect the SHM would be. So we will assume that the ball does not move beyond say 10 degrees. And as I said, smaller the angle, the better it is. Well if this is the case then the cos of the angle will be close to one. How close to 1? Well, at 5 degree, cos of 5 is 0.996, Infact event at 10 degrees, it is 0.985 which actually is just 1.5% off 1
2. As a consequence of this small angle assumption, what you will find is that the acceleration of the ball in the vertical direction will be very small. This is true since the displacement of the ball in vertical direction, for low theta values, will be very small. In fact at 5 degree displacement, the y displacement will be just 4.5% of the x displacement.

Well if these are the 2 conditions then equation 2 becomes

0 – T – mg or t = mg
If we substitute this into equation 1, what we get is ma = -mgx/l or a = -g/l x
Which is nothing but equation of SHM. And it is so because if you recall from the earlier lesson, the condition for SHM is that the displacement and acceleration should be opposite in direction at all times and the two variables need to be connected by a certain constant which was k/m in case of spring.
So you can see that this equation is also satisfying conditions of a SHM motion.
Here the ….
You can see that these expressions do not involve mass m. The reason is that the restoring force is proportional to m. In other words, more the mass, more the restoring force. Hnece when you write the equation F = ma, mass cancels on both sides thus making dependence on mass as zero. Thus the time period is dependent on length L and this seems quite intuitive, you know that a longer string takes more time to swing one arc. It also depend on gravity g but inversely. That is lower the g, higher the time period T. Thus if you are in outer space where say g is close to zero, the mass will take a very long time to swing through an arc.

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