# Why is the method I'm using incorrect to find out the time taken for the particle to complete 5/8 oscillations?

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.

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?

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

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.