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I know that the energy of SHM is given by

$$E=\frac 12 kA^2$$

So which of these is more appropriate to say?

  1. Energy is increases because amplitude is increased.

  2. Amplitude is increased because energy is increased.

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It depends on the context of the question you're asked to solve using this formula. The formula you've written is that of the total energy of a simple harmonic oscillator, such as a block attached to a spring.

enter image description here

At the maximum displacement, there is no kinetic energy and the potential energy is hence given by $\frac 12 kA^2$. Note that from this, $E \propto A^2$.

Now, if we pull the block so that its amplitude increases, we are giving energy to the system. This means that the energy and the amplitude both increase.

Do note that we can't say one happened because of another, it really depends on the case you're looking at. (A parallel you can draw is, in $\vec F=m \vec a$, if you increase the force applied, the body's acceleration will increase, and if you note an increase in the body's acceleration, of course the force must be increasing.)

In most cases, however, we usually are able to easily change the amplitude manually, because of which the energy of the system changes. Changing the amplitude is the way in which we increase the energy. In that sense, the energy increase happened because we changed the amplitude.

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  • $\begingroup$ Thank you for your reply. However I still have one doubt. Let me take the same example of F=MA for convenience. As you mentioned if Force increases that causes the acceleration to increase. But I don't agree with the reverse of that. If acceleration increases, doesn't that mean that the force has increased? Because acceleration is a result of the Force, but Force is not the result of acceleration-it causes it. So is there is a similar analogy for the SHM? $\endgroup$ – Vamsi Krishna May 1 at 6:42
  • $\begingroup$ And thank you for the nice animation $\endgroup$ – Vamsi Krishna May 1 at 6:43
  • $\begingroup$ Mathematically, there is no difference. However, yes, like you said, typically, we can only increase force. Acceleration is a result of that. Maybe mass is a better example, if we increase the mass, can we say acceleration decreased? It's just the math, there isn't a a need for this confusion. Generally, we increase amplitude. So yes, it is more correct to say that energy increases because of an amplitude change. Still, it's possible we have some system where if we supply some energy (say, heat), the amplitude will also change. In general though, usually it's the analogy you mentioned. $\endgroup$ – wavion May 1 at 6:50
  • $\begingroup$ Thanks it helped. But just another doubt. The amplitude can not just increase on its own right? If amplitude has increased, doesn't that mean that mean that work has been done for the same? So doesn't it imply that amplitude increases because energy increased? $\endgroup$ – Vamsi Krishna May 1 at 7:15
  • $\begingroup$ The amplitude cannot increase on its own. Let us say that we displace the block extra distance $x$ and the new amplitude is $A+x$. This means that the energy increased because we increased the amplitude. I think this might be an English issue: 'because' implies a cause-effect relation. The energy increases because we increase the amplitude. Put differently, energy increase is the result of the amplitude increase. So no, it does not imply that amplitude increases because energy increases. Energy increasing is kind of a 'side-effect' $\endgroup$ – wavion May 1 at 7:32
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There is no 'because' in an equation. $a=b$ does not imply $a \to b$ or $b \leftarrow a$.

If you increase the energy of an oscillator, you will increase its amplitude. If you increase its amplitude you will increase its energy.

This is brought out neatly in our use of Ohm's law. $V=IR$ is usually read and used as 'If you apply $V$ volts across $R$ ohms you get $I$ amps.' But sometimes (for example in a potential divider) you use it as 'There is a current of $I$ amps flowing through $R$ ohms so that gives $V$ volts".

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  • $\begingroup$ Thank you roger $\endgroup$ – Vamsi Krishna May 1 at 17:10

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