1. ## SHM

given that x=120 sin (pi/12)t , what is the first time when the particle is (i) 30m to the right of the origin and (ii) 30m to the left of the origin?

Im thinking

30=120 sin (pi/12)t

1/4=sin (pi/12)t

t=(1/4) / sin (pi/12)

t=0.96592..... is this correct?

I don't know how to think about the second part when x=-30. Is it more complicated or do I just solve it as above?

Thank you

2. Originally Posted by slaypullingcat
given that x=120 sin (pi/12)t , what is the first time when the particle is (i) 30m to the right of the origin and (ii) 30m to the left of the origin?

Im thinking

30=120 sin (pi/12)t

1/4=sin (pi/12)t

t=(1/4) / sin (pi/12)

t=0.96592..... is this correct?

I don't know how to think about the second part when x=-30. Is it more complicated or do I just solve it as above?

Thank you
No, it is wrong.

$30 = 120 \sin \left(\frac{\pi}{12}t\right)$

$\frac{1}{4}=\sin \left(\frac{\pi}{12}t\right)$

$0.25268 = \frac{\pi}{12}t$

$t=0.25268 \times\frac{12}{\pi}$

$t=0.9652$