# Reciprocating piston problem, can you solve it?

• Mar 20th 2014, 08:33 AM
TehSwedishBeast
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
• Mar 20th 2014, 08:40 AM
SlipEternal
Re: Reciprocating piston problem, can you solve it?
What are you having trouble with? Do you know for what values of $\theta$, $\cos \theta = 0$?
• Mar 21st 2014, 09:08 AM
TehSwedishBeast
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
• Mar 21st 2014, 09:51 AM
romsek
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
• Mar 21st 2014, 03:41 PM
SlipEternal
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.
• Mar 23rd 2014, 04:02 AM
TehSwedishBeast
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?
• Mar 23rd 2014, 04:07 AM
romsek
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.