Solving for t in SHM (complex solution)

[mouser]

Hey guys, this is my first post, I was hoping you all could offer some advice. I'm facing a problem involving free undamped SHM. Everything is working out well so far but I ran into a problem when trying to find t. Here's what I have:

x(t) = -2/3cos10t + 1/2sin10t

Now, if I want to find the times when x(t) = 0... how would I go about that? In a similar problem that only involved a single term, I was able to use cos^-1 to solve for t, but in this case would that still hold? Setting x(t) to 0 I get:

0 = -2/3cos10t + 1/2sin10t

But now I'm stuck.

Any help would be greatly appreciated!

 

[tiny-tim]

Hi mouser! Welcome to PF!

0 = -2/3cos10t + 1/2sin10t

Hint: how might you re-arrange this equation?

[size=-2](if you're happy, don't forget to mark thread "solved"!)[/size]​

 

[mouser]

Thanks!

If I rearrange it to be:

2/3cos10t = 1/2sin10t

4/3cot10t = 1

3/4tan10t = 1

tan10t = 4/3

t = (tan^-1(4/3))/10

The answer is incorrect. Did I make a mistake in that algebra? Thank you for your help!

 

[tiny-tim]

t = (tan^-1(4/3))/10

Looks ok to me!

Except you only have one solution … there should be infinitely many … what are the others?

(What were you actually asked for? Was it the times, or the period?)

 

[mouser]

The question was "at what time does the mass pass through the equilibrium position heading downward for the second time?"

I just figured out where I went wrong. The answer I got previously was only partially correct. For the correct answer I added the value for one full period and it checks out! eureka!

Thanks tiny-tim!