Reciprocating piston problem, can you solve it?

The displacement, in metres, of a reciprocating piston, from its point of equilibrium (the mean) is given by the following function: x= 3 cos( 4t + π

/2) where t is time in seconds.

Starting from rest, find the first three times at which the piston passes through the mean position at x=0

Re: Reciprocating piston problem, can you solve it?

What are you having trouble with? Do you know for what values of $\theta$, $\cos \theta = 0$?

Re: Reciprocating piston problem, can you solve it?

I don't know where to start with the question and no I don't know for what values of theta or cos theta=o

Re: Reciprocating piston problem, can you solve it?

Quote:

Originally Posted by

**TehSwedishBeast** I don't know where to start with the question and no I don't know for what values of theta or cos theta=o

your told that the displacement $x$ is given by

$x= 3 \cos(4t + \dfrac{\pi}{2})$

and you need to find when $x=0$ so just set this expression equal to zero and solve it.

$0= 3 \cos(4t + \dfrac{\pi}{2})$

$0=\cos(4t + \dfrac{\pi}{2})$

Now you have to know that $\cos(\theta)=0$ when $\theta=\pi (n+\dfrac{1}{2})$ for n being some integer. Set n=0,1,2 to find the first 3 times $x=0$

so for example with n=0

$4t + \dfrac{\pi}{2} = \dfrac{\pi}{2}$

$4t=0$

$t=0$

you can solve for the t's corresponding to n=1 and n=2

Re: Reciprocating piston problem, can you solve it?

Quote:

Originally Posted by

**TehSwedishBeast** I don't know where to start with the question and no I don't know for what values of theta or cos theta=o

romsek gave you the answer, but the only way you would not know for what values of $\theta$ give $\cos \theta = 0$ is if you never studied trigonometry. I would recommend holding off on answering the question until after you study trigonometry.

Re: Reciprocating piston problem, can you solve it?

I understand now so for n=1 would it be:

4t + π/2= π/2

4t= 1

t= 0.25

and for n=2 would it be:

4t + π/2 = π/2

4t= 2

t=0.5

Am I right with that?

Re: Reciprocating piston problem, can you solve it?

Quote:

Originally Posted by

**TehSwedishBeast** I understand now so for n=1 would it be:

4t + π/2= π/2

4t= 1

t= 0.25

and for n=2 would it be:

4t + π/2 = π/2

4t= 2

t=0.5

Am I right with that?

no.

for n=1 we have

$4t + \dfrac{\pi}{2}=\pi(1+\frac{1}{2})=\dfrac{3\pi}{2}$

$4t = \pi$

$t=\dfrac{\pi}{4}$

for n=2 the right hand side is $\pi(2+\frac{1}{2})=\dfrac{5\pi}{2}$

Now solve this for t.