Show that the tension in a simple pendulum is of the form $$T(t)= T_0 + T_2 \cos(\omega t)$$ and find $T_0$ and $T_2$ in terms of the mass $m$ of the bob, the amplitude of the oscillation $θ_0$ and acceleration due to gravity, $g$.

I understand how to resolve the forces and have the correct answer, but I do not know why it is correct. I resolved such that:

$$T-mg\cos\theta = \frac{mv^2}l$$ and worked from there to get the answer. But why is this resolving correct? Isn't the centripetal acceleration in a perpendicular direction to the components I resolved?


closed as off-topic by John Rennie, Thomas Fritsch, eranreches, GiorgioP, glS Jun 10 at 16:15

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  • $\begingroup$ When you resolved the components of the forces, did you chose the radial component? That is the component perpendicular to the velocity. Why would you want to be perpendicular to the velocity-based perpendicular component? I think you confused yourself by going one step too far. $\endgroup$ – Bill N Jun 8 at 12:57

Centripetal acceleration is in radial direction and towards the centre. The resultant of component of mg in radial direction and tension provides the required centripetal force. Tension is already along the radial direction. The other(tangential) component of mg provides Tangential acceleration which speeds up or down the pendulum. In case of analysing a system where centripetal forces are involved, it is better to resolve into tangential and radial components.


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