What Is the Minimum Time for SHM Particle to Travel Between Two Points?

[blue_leaf77]

Because trig functions have always angles (either in radians or in degrees) as an argument / input . Right?

Yes.

What other quantities can become argument of trig functions?

As far as I know, no other.

 

[gracy]

Thank you so so much my friend @blue_leaf77 . God bless you

 

[blue_leaf77]

Thank you so so much my friend @blue_leaf77 . God bless you

Glad to help you.

 

[gracy]

One more question. We can't take directly y=25 cm (12.5+12.5)cm i.e distance between points (points of interest) a & b

because "y" in the equation $y$=$a$ sin $w$$t$ is distance from mean position . I also tried to calculate it
25=25 sin (120 t)
1=sin (120 t)
(120 t)=90
t=90/120
t=0.75 s
Look, I got wrong answer, I should've got 0.5 s. This proves what I just wrote above. Right?

 

[blue_leaf77]

because "y" in the equation yyy=aaa sin wwwttt is distance from mean position .

Yes, that exactly the reason.

25=25 sin (120 t)
1=sin (120 t)
(120 t)=90
t=90/120
t=0.75 s

You are basically calculating the time needed from the mean position to the point of farthest displacement - the amplitude. That's why what you got is a quarter of the period.