In SHM, acceleration is always in the opposite direction of the oscillating particle.But, in the case of a pendulum, when the bob is, at first, set free, it's displacement is towards the acceleration before reaching the equilibrium position. So why is the acceleration of the oscillating bob in the direction of its displacement?

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    $\begingroup$ Sorry but I don't understand your question. $\endgroup$ – Suzu Hirose Oct 27 '16 at 5:00
  • $\begingroup$ i asked that if the acceleration is in the opposite direction of the displacement in shm, but in case of pendulum i dont understand the case when the bob first move towards the pendulum how the acceleration is in the opposite side of the displacement. $\endgroup$ – Hemanta Paul Oct 27 '16 at 5:04
  • $\begingroup$ I don't think it is in the opposite direction. If you pull it right, it accelerates left. $\endgroup$ – Suzu Hirose Oct 27 '16 at 5:09
  • $\begingroup$ no thats ok. i pull it right it accelerates left but when the acceleration is toward the equilibrium position i.e. towards left it's displacement is towards the acceleration. $\endgroup$ – Hemanta Paul Oct 27 '16 at 5:15
  • $\begingroup$ In a pendulum , when you let go of the bob from the, let's say, left extremum, it swings, covers a distance and moves to the right extremum. But it also comes back to its initial position(of course ignoring the damping), so net displacement is zero for one whole cycle, as it has not covered any distance in whole cycle. The restoring force, however, acts on the bob to bring it back to a position midway in the path(equilibrium position). So since it acts on the bob to bring it back(restoring force), the acceleration is opposite to its initial displacement from the midway position. Hence SHM. $\endgroup$ – Naveen Balaji Oct 27 '16 at 5:20

You are right in saying that in SHM (Simple Harmonic Motion) the acceleration is opposite to the displacement.

Suppose you have a pendulum of some mass M hanging from a string as shown below.

enter image description here

Here A is the mean position of the bob of the pendelum.

When the bob of the pendelum was at position A then the force of gravity was acting downwards and there was so sideways component of force on the pendelum.

When you displaced it towards the position "B" from "A" the displacement of the particle is from left to right (A to B or from mean position to extreme position)

But The force of gravity still acts downwards and when you break it into its components then one of the components mgsin(θ) acts in a direction that tries to bring the bob back to mean position i.e Force on the bob acts from right to left or from "B" to "A"

Since from newton's second law

F = ma


a = F/m

From the above law we can see that if Force acts from right to left then so does the acceleration.

Thus acceleration of the particle is towards the mean position (right to left or from "B" to "A") but your displacement was away from the mean position i.e from "A" to "B"

Hence acceleration and displacements have opposite directions. I hope I've cleared your confusion.

P.S If by displacement you mean where the particle wants to move now then Its clear that it wants to move towards A and the acceleration is also towards A, So in this sense they are in same direction But then the particle passes position A and it going to the other extreme end then again component of gravity acts towards A and so again acceleration and displacement are in opposite directions

  • $\begingroup$ so, at first when the particle moves towards 'A' i.e. the equilibrium position it's acceleration is towards the displacement,then when it crosses the equilibrium position its acceleration acts opposite the displacemnt. $\endgroup$ – Hemanta Paul Oct 27 '16 at 6:40
  • $\begingroup$ Yes. Because again one component of gravity mg sin(θ) acts towards "A" $\endgroup$ – Kashish Oct 27 '16 at 6:57
  • $\begingroup$ then what you will say about this formula f=-kx $\endgroup$ – Hemanta Paul Oct 27 '16 at 7:38
  • $\begingroup$ x=displacement k is the proportionality constant. Here acceleration is always negative.That means when the particle is moving towards A the acceleration is negative i.e opposite side and how this is possible. $\endgroup$ – Hemanta Paul Oct 27 '16 at 7:45
  • $\begingroup$ F = -kx is perfectly right equation of SHM. Don't forget it is a vector equation. minus sign denotes only direction. You might want to take a look at vector notation and how they are used. This is a totally different topic. You might use the following video link to have a quick look at vectors $\endgroup$ – Kashish Oct 27 '16 at 22:58

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