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bayley86 · Dec 9, 2010, 12:05 PM

A piston performs reciprocal motion in a straight line which can be modeled as SHM. Its velocity is 16 m/s when the displacement is 80 mm from its midpoint, and 4 m/s when the displacement is 160 mm from mid-position. Determine the frequency and amplitude of the motion and the acceleration when displacement is 120 mm from mid position

Unknown008 · Dec 9, 2010, 12:23 PM

The velocity is linked to the displacement as by the formula

$v = \omega \sqrt{x_o\ ^2 - x^2}$

Where v is the velocity at any point x from the mid-position,
omega the angular velocity,
xo the amplitude
x the displacement from the mid-position.

Acceleration is then given by:
$a = -\omega^2 x$

bayley86 · Dec 14, 2010, 10:30 AM

I still don't understand this could you please explain what I have to do abit more thanks

ebaines · Dec 14, 2010, 11:33 AM

Using the equations that Unk gave you:

$ v = \omega \sqrt{x_0^2 - x^2} $

where $x_0$ is the amplitude of the SHM, and $\omega$ is the rotational veocity in radians/second.

You have been given two data points:

v = 16 m/s when x = 0.08m;
v = 4 m/s when x = 0.16m

Note that I changed the dimensions to be consistenty in meters. Now substitute those values into the above equation, to give you two equations in two unknowns:

$ 16m/s = \omega \sqrt{x_0^2-0.08m^2} \\ 4m/s = \omega \sqrt{x_0^2 - 0.16m^2} $

Solve this for $\omega$ and $x_0$ . Then the frequency of the system is: $f = \frac {\omega}{2 \pi} $ and the acceleraton when x = 0.12m is: $a= -\omega^2 \cdot 0.12m $ in meters per second squared.

bayley86 · Dec 14, 2010, 12:08 PM

if I divide both these equations by each other dose that illiminate w and the squaring but I still can't figure out x

bayley86 · Dec 14, 2010, 12:20 PM

is x 62.5

ebaines · Dec 14, 2010, 12:21 PM

Originally Posted by bayley86

if i divide both these equations by each other dose that illiminate w and the squaring but i still can't figure out x

You're on the right track:

$ \frac {16} 4 = \frac {\omega \sqrt{ x_0^2 - 0.08^2} } {\omega \sqrt{ x_0^2 - 0.16^2} } \\ \\ 4 = \sqrt {\frac {x_0^2 - 0.08^2}{x_0^2 - 0.16^2}} $

Square both sides:

$ 16 = \frac {x_0^2 - 0.08^2}{x_0^2 - 0.16^2} $

Multiply through by the denominatoir:

$ 16 (x_0^2 - 0.16^2) = x_0^2 - 0.08^2} $

Rearrange to get x_0 by itself:

$ 16x_0^2 - x_0^2 = 16 \times 0.16^2 - 0.08^2 \\ x_0^2 = \frac {16 \times 0.16^2 - 0.08^2} {15} \\ x_0 = \sqrt {\frac {16 \times 0.16^2 - 0.08^2} {15}} $

ebaines · Dec 14, 2010, 12:23 PM

Originally Posted by bayley86

is x 62.5

62.5 what? Millimeters? And do you mean x_0? Obviously this is not right since you are given data about what happens when x = 80mm or x = 160 mm. Clearly x_0 must be greater than 160 mm.

bayley86 · Dec 14, 2010, 01:03 PM

from that equations I got 0.04233 so is x 423.3 mm

ebaines · Dec 14, 2010, 01:13 PM

Originally Posted by bayley86

from that equations i got 0.04233 so is x 423.3 mm

Nope. I think when you did the calculation you may not have reaized that the denominator is inside the square root. In other words you took the square root of the numerator and then divided by 15, as opposed to dividing the numerator by 15 and then taking the square root.

Also - if the answer had turned out to be 0.04233m that would be 42.3mm, not 423.3 mm. Watch your units!

bayley86 · Dec 14, 2010, 01:18 PM

is x 639.6 mm

ebaines · Dec 14, 2010, 01:22 PM

Originally Posted by bayley86

is x 639.6 mm

No.

bayley86 · Dec 14, 2010, 01:29 PM

Do you no what I'm doing wrong

bayley86 · Dec 14, 2010, 01:39 PM

is x 163.9

ebaines · Dec 14, 2010, 01:44 PM

Originally Posted by bayley86

is x 163.9

Bingo! But the answer should properly be rounded to 164mm, since that's the level of accuracy of the data you were given.

bayley86 · Dec 14, 2010, 01:49 PM

I think I need the equation for angular velocity to find the frequency do you no what this is equation is

ebaines · Dec 14, 2010, 01:57 PM

Go back to post #4. There were two equations in two unknowns: $\omega$ and $x_o$ . You have now solved for $x_o$ . So you can use that knowledge to find $\omega$ , by plugging the value for $x_o$ into either one of those equations and solving for $\omega$ .

bayley86 · Dec 14, 2010, 02:38 PM

do I have to rearrange the equation

ebaines · Dec 14, 2010, 02:55 PM

Originally Posted by bayley86

do i have to rearrange the equation

Sure. For example if you use this one:

$ 16 \frac m s = \omega \sqrt{x_0^2-(0.08m)^2} $

Rearange to get $\omega$ by itself, then plug in the value for $x_o$ , and you've got it.

bayley86 · Dec 14, 2010, 03:11 PM

dose w = 106rad/s