Question

find the period of the function f(t) =[sin(10*pi*t)]^n where ^ signifies 'to the power' when n is odd and even.

Answer 1

Period of \(\Large \sin(10\pi t)\) is \(\Large \frac{2\pi}{10\pi} = \frac 15\)
Period of \(\Large [\sin(10\pi t)]^n\) when n is odd should also be \(\Large \frac 15\).
Period of \(\Large [\sin(10\pi t)]^n\) when n is even should be \(\Large \frac 25\).

Answer 2

nope... for even 1/10 .. i think .....

Answer 3

Oh, yes. Frequency is doubled. Period should be half.

Answer 4

right @aum

Answer 5

There is a reason to that.

When n is even. All the negative values of sin are no longer negative and hence, the function repeats faster than when it is odd. The zeroes remain the same.

When n is odd. The negative values remain negative, and the zeroes remain the same. So the function repeats like it usually should.

When n is even = Period = \(\frac{1}{10}\)
When n is odd = Period = \(\frac{1}{5}\)

Do you understand this? :)

Answer 6

yaah