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The following diagram shows the direction of the acceleration of a pendulum at different states. image1 Also, consider the below diagram. It shows the state before the pendulum starts simple harmonic motion.

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In this situation, a horizontal force is exerted on the bob in addition to the tension by the string and its weight and also it is in rest. According to the theorem of the parallelogram of forces, the resultant of $T$ and $mg$ acts horizontally along the opposite direction of $F$. Just after $F$ is removed, the net force on the bob should act along the horizontal axis. According to $\vec F=m\vec a$, the initial acceleration should be along the horizontal axis. But this idea contrasts with the first diagram mentioned above.

In my opinion, if the bob moves a little bit ahead horizontally, the string will loosen and there will be no tension. So it will gain a downward acceleration caused by its weight. When it tries to go down the string tightens again and tension arises. So the direction of the net force is changed again with the direction of acceleration. This is unnoticeable in comparison with time. But I am still not sure about this. Please help me for a better solution.

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    $\begingroup$ The key point it that the simple analysis that provides the result in your top figure assumes a rigid string (constant length); perhaps a rigid rod is a better description. The tension is as necessary to prevent radial motion of the mass at the end of the string (rod), and the tension changes as necessary to fulfill this requirement. Answers provided by others discuss this in more detail. $\endgroup$
    – John Darby
    Jul 10, 2021 at 23:51

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The force F, which you have drawn horizontally, could actually have been drawn in a number of different directions, including the direction opposing the starting acceleration.

The tension in the string changes and the required size of F changes as the direction of F is adjusted. If F points in the direction opposing the initial acceleration then the size of F will be exactly equal to the initial accelerating force on the bob.

At all times the tension in the string will automatically adjust to keep the total force zero. When F ceases the tension changes to the appropriate value and the mass begins to move.

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  • $\begingroup$ Then, is this statement false?:"According to the theorem of the parallelogram of forces, the resultant of T and mg acts horizontally along the opposite direction of F. Just after F is removed, the net force on the bob should act along the horizontal axis. According to $\vec F=m \vec a$, the initial acceleration should be along the horizontal axis." $\endgroup$
    – ACB
    Jul 9, 2021 at 13:25
  • $\begingroup$ The statement about parallelogram of forces is correct. However T changes as soon as F is removed. If you want to see the horizontal movement then use a spring instead of the pendulum string or rod, so that you can see what happens as the tension changes. In this case your statement would be completely correct. The apparatus won't be a simple pendulum any more though. $\endgroup$
    – Peter
    Jul 9, 2021 at 14:48
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You explained it almost correctly in your last paragraph. I'm elaborating it to explain exactly what happens when $F$ is removed.

The moment the horizontal $F$ is removed, tension $T$ changes to $T'$ (we'll see later what this $T'$ is).

The downward force of $mg$ is always there and the vertical component of $T$ ($T_v$) was taking care of that. The horizontal component of $T$ ($T_h$) was taking care of $F$. So, initially we had:$$ \begin{align} T_h &= F \\ T_v &= mg \end{align} $$

Right after $F$ was removed, $T$ changes to $T'$. This happens because the bob starts to move under the influence of $T_h$ which immediately loosens the string.

Like $T$, the new tension, $T'$, is there only because of the fixed-length-property of the string fixed at one end (barring very little stretchability that generates the tension), i.e. because of the constraints that the length of the string cannot change and the other end of the string cannot move in the direction of the force. Hence, $T'$ automatically becomes equal to whatever force is trying to lengthen the string, which is $mg\cos{\theta}$. So, we have this now:$$ T' = mg\cos{\theta} $$ After all this, there still remains a force that is not counter-balanced - $mg\sin{\theta}$. Hence, this is the force $(mg\sin{\theta})$ that causes the acceleration of the bob. At any point in the journey of the bob, $T'$ keeps adjusting itself by becoming equal to $mg\cos{\theta}$ simply because the length of the string can't change. And, hence there is always $mg\sin{\theta}$ left to cause acceleration.

Since $\theta$ keeps on changing, both the direction and magnitude of the net acceleration keep on changing during the journey.

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    $\begingroup$ Your explanation seems better. But I have a doubt in this statement:"Hence, T′ automatically becomes equal to whatever force is trying to lengthen the string". I am confused with this when thinking of an accelerating system.e.g.atwood maching. I know that these are two different situations. But I am asking whether this statement always true. If 'no', there is another question. $\endgroup$
    – ACB
    Jul 10, 2021 at 6:29
  • $\begingroup$ Valid point. This statement - "Hence, T′ automatically becomes equal to whatever force is trying to lengthen the string" - is only true when the other end of the string is not allowed to move at all in the direction of the force on the string. In this case, the string will break, stretch or have its tension equal to the pulling force. If this end is allowed to move in the direction of the force, tension will be different from the pulling force (as happens in Atwood machine). $\endgroup$
    – manisar
    Jul 10, 2021 at 16:12
  • $\begingroup$ It is now clear and problem solved. Thank you. $\endgroup$
    – ACB
    Jul 10, 2021 at 16:34
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You are comparing two different situations.

When the horizontal force $F$ is applied to the bob it stretches the string so that there is tension $T=\frac{mg}{\cos \theta}$ in the string, and thus the vertical component of $T$ is equal and opposite to the weight of the bob. By equating the horizontal component of $T$ with $F$ you can find the angle $\theta$. When the bob is released from this position it does indeed initially accelerate horizontally because the horizontal component of $T$ is now not balanced by $F$. As the bob moves horizontally it reduces the tension in the string.

But if the the bob is swinging freely, then at the extreme point of each swing we have $T=mg \cos \theta$ because the bob is accelerating tangentially, so the net radial force on it is zero.

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    $\begingroup$ The second paragraph is only true if the amplitude of the swing is exactly +/- 90 degrees. $\endgroup$
    – Ben51
    Jul 9, 2021 at 15:35
  • $\begingroup$ @Ben51 Yes indeed - my bad. I have rewritten the second paragraph. $\endgroup$
    – gandalf61
    Jul 9, 2021 at 16:49
  • $\begingroup$ Thank you @gandalf61. As you say, I am comparing two different situations. Also please refer to the following question:physics.stackexchange.com/questions/649985/… . I am sure it is the situation mentioned in your first paragraph. And I can't understand why my answer(as explained in this post) was wrong there.(I remember you also answered that question and I guess you can view my deleted answer too.) Any idea? $\endgroup$
    – ACB
    Jul 11, 2021 at 3:02

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