Question

For the simple harmonic motion equation d=5sin((pi/4)t), what is the maximum displacement from the equilibrium position?

Answer 1

\(\color{black}{-1\le \sin\theta \le 1}\)

Answer 2

In general, the maximum value of \(\color{black}{ \sin(\beta \theta) }\) for all \(\color{black}{ \beta }\) is \(\color{black}{ \sin(\beta \theta) =1}\).
So, for \(\color{black}{ \alpha >0 }\), the maximum of \(\alpha \color{black}{ \sin(\beta \theta) }\) is \( \color{black}{ \alpha \times 1 = \alpha }\).

Answer 3

so the answer would be 5 or -5

Answer 4

Displacement does denote a vector quantity, but I suspect that your question is looking for the maximum distance that you can get away from equilibrium. (Although this last assertion, I wouldn't bet my life on.)

Answer 5

this is my second time getting this question . the first time i put 5 and i wondering if i put that again

Answer 6

You didn't know whether or not you got it correctly the first time when you put 5 as the answer?

Answer 7

don't (not "didn't)

Answer 8

Well, if it was me, I would say that the maximum displacement from the equilibrium is 5. (Without -5).

Answer 9

it's 5 thank you

Answer 10

yw