Period

[fr33pl4gu3]

f(x) = - sin(x/5 - pi)

the period for this is -pi correct, since, k taken out it is 1, and by divide -pi into 1, the answer will be -pi only, correct??

[mgb_phys]

I don't think you can have a negative period!
To find the period pick a value of x (say zero) and then work out what is the next value of x that will give the same value of f().
Does adding the same constant to both values of X do anything?

[fr33pl4gu3]

Does it gives the same pi??

[fr33pl4gu3]

So, the answer is "pi"??

[mgb_phys]

The period of sin() is pi ie sin(0)=sin(pi)

But in terms of X, since you divide X by 2 what value of X do you need for the function to give the same value as for x=0?

[fr33pl4gu3]

What is the exact period of f(x) (in radians)? (Recall p radians is equivalent to 180°; p is obtained by entering: Pi or pi.)

What does the question actually wants anyway?? how to calculate period in radians.

[fr33pl4gu3]

zero??

[Avodyne]

The period of sin() is pi ie sin(0)=sin(pi)

This is not correct. A function f(x) is periodic with period p if
f(x+p)=f(x)
for all real x. For \sin(x), this is true for p=2\pi, and not true for p=\pi, because \sin(x+\pi)=-\sin(x), not \sin(x).

[fr33pl4gu3]

yes, and the answer is 10pi, by getting the 2pi divide 1/5 to get the answer. I get confuse that i need to multiply by 5 so to get b=5, but when i try 1/5, it's done.

[Avodyne]

yes, and the answer is 10pi

Correct!

[Astronuc]

in this function, f(x) = - sin(x/5 - pi), the -pi would be a phase shift.

In general for sin (wt + q), w = 2pi f = 2 pi/T, where T is the period, and q is the phase shift.

[HallsofIvy]

The period of sin() is pi ie sin(0)=sin(pi)

But in terms of X, since you divide X by 2 what value of X do you need for the function to give the same value as for x=0?

No, the period of sin(x) is 2pi.

fr33pl4gu3. your function is -sin(x/5- pi). When x/5- pi= 0, what is x? When x/5- pi= 2pi, what is x? The difference between those is the period.