We use phasor diagram for SHM so when the phasor travels $2\pi$ radians it's projection on the y axis would have completed one oscillation. It says in the question that the particle completes $5/8$ of its oscillation.
So $2\pi\to1$ oscillation.

$X\to\frac58$ oscillation which gives $$X=\frac54\pi=\pi+\frac{\pi}{4}$$ This is what I used.

Question: Question (1)

Given solution: Given Solution

My method: My Method

There was a similar question which asked to find out the displacement of a particle from mean position at time $T/8$ given that the time period of the particle was $T$ and the method I used above gave me the correct answer. Where am I going wrong?

  • $\begingroup$ Hello! It is preferable to type out screenshots; for formulae, one can use MathJax. Thanks! $\endgroup$ – Jonas May 27 at 16:44
  • $\begingroup$ @Jonas idk, when the problem is a train wreck like this, it's good to see the original. $\endgroup$ – JEB May 27 at 16:50

It depends whether '1/8 of an oscillation' refers to 1/8 of the phase corresponding to a complete oscillation, or 1/8 of the distance. They are different because the particle speed varies. You use the former, the answer uses the latter.

After 30 degrees (as shown) the particle has travelled $A\sin(30)=A/2$ cm, which is halfway to the full A cm and thus 1/8 of the distance travelled during the cycle. They call that 1/8 of a cycle, you call it 1/12.

I would agree with your usage, and I expect most others would too. But that's where the discrepancy lies.


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