Question

for the simple harmonic motion equation d=9cos((pi/2)t) what is the period

Answer 1

For cos(t), t going from 0 to 2pi is one period. If I change the function to cos(2t), then t going from 0 to pi makes (2t) go from 0 to 2pi. So now the period is just pi.

Answer 2

is t in the denominator or the numerator?

Answer 3

d=9cos((pi/2)t)

Answer 4

Doesn't it have to be the numerator? Otherwise there would be no discernible period.

Answer 5

right, but I think your answer is wrong SM...
for functions of the form\[A\cos(Bx+C)\]the period is given by
\[T=\frac{2\pi}{|B|}\]

Answer 6

looks like T = 4 ???

Answer 7

the standard equation of SHM might be
\[ y = A\cos(\omega t + \phi)\]

And omega might be your angular frequency.

you may calculate using analogy between two equations.

Answer 8

and \[\omega = \frac{2\pi}{T}\]

where is T is your time period.

Answer 9

No, you're right, Turing. I was just giving an example of how it's done. I wasn't working the problem for him/her.

Answer 10

scooby doovey do .... just give some clue.

Answer 11

lol^

Answer 12

\[\omega t=(2\pi f)t=(\pi /2)t\]
\[2\pi f=\pi /2\]
\[f=(\pi /2)\times(1/2\pi)=(1/4)Hz\]
Period=1/f=4seconds

Answer 13

yes you are right.