Question

For the simple harmonic motion equation d=5sin (pi/4^+), what is the period?

Answer 1

8

Explanation:

A

P

E

X

Answer 2

the period is 8

Explanation

the function sin has the form

`\begin{gathered} y=Asin(B(x+c))+D \\ where \\ Period=(2\pi)/(B) \end{gathered}`

so

Step 1

a) identify B in the given function

given

`d=5\text{ sin\lparen}(\pi)/(4)t)`

hence

`\begin{gathered} (\pi)/(4)t\Rightarrow B(t+c) \\ so \\ c=0 \\ (\pi)/(4)t=Bt \\ therefore \\ B=(\pi)/(4) \end{gathered}`

b) now, replace in the formula to find teh period

`\begin{gathered} Per\imaginaryI od=(2\pi)/(B) \\ Period=(2\pi)/((\pi)/(4))=(2\pi *4)/(1*\pi)=(8\pi)/(\pi)=8 \\ so \\ Period=8 \end{gathered}`

therefore, the period is 8

I hope this helps you