(Not so) Simple Harmonic Motion
by XXXXXXX and Jedi Knight Dustin Glazier, February 1999
Table of Contents
Here's the quantities you can know:
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X Displacement
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t Time
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T period
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f frequency
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Vo Initial velocity
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Va Average Velocity
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V Final Velocity
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a Acceleration
-
k spring constant
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m mass
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F Force
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g gravity
New Quantities
And here are the formulas that we have so far:
Defining
El Construction Zone
Formulas
-
El Construction Zone
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General Problem Solving Strategy:
-
Read the problem.
-
Go through the problem and figure out what is given or implied
Make a list, and identify the quantities you know.
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Find any formula that will allow you to calculate
anything that you don't know, and apply it.
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Add what you just found in the last step to your list of knowns.
-
Check to see if you have found the answer. If not, repeat the
previous two steps until you are done.
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Example problem 1
El Construction Zone
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Example problem 2
El Construction Zone
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Sample Problems
el solvedo bi el Jedi Knight Dustin Glazier
The answers to each problem follow it in parentheses. They also link to a solution to
the problem. Try the problem, check your answer, and go to the solution if you do not
understand.
1.
A spring has a restoring force of 320 N when it is stretched 22.1 cm.
What is the spring's constant k in N/m?
(14.5 N/m)
2.
A very stiff spring has a k = 24350 N/m. What is its restoring force
when you distort it 1.00 cm?
(234.5 N)
3.
Another totally different spring has a spring constant of 249 N/m.
How far will it distort when you hang a 5.400 Kg mass from it?
(.213 m)
4.
A spring has a constant of 34 N/m, and a 3.00 kg mass hangs from it.
What would be the period of motion of the mass on the spring?
(21.2 seconds) Go back to: Table
of Contents
5.
You need a period of exactly 1.00 seconds, and you have a mass of 230 grams.
What must be the spring constant of the spring?
(.0058 N/m)
6.
You need a period of exactly 1.00 seconds, and you have a spring constant
of 120. N/m. What must be the mass on the spring (in grams)? (4700000
grams)
7.
Joe takes an unknown spring and hangs a 5.000 Kg mass on
it. The weight of the mass stretches the spring 230.3 cm. What
would you expect the period of the mass to be if it were set in motion?
(41
seconds)
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Solutions to Sample Problems
1.
A spring has a restoring force of 320 N when it is stretched 22.1 cm.
What is the spring's constant k in N/m?
(14.5 N/m)
Here is what you start with:
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X = 22.1 cm
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F = 320 N
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k = ?
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m = ?
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g = 9.8 m/s/s
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T = ?
Well, since we're just starting out, this seems quite easy. We take the
formula F = kx, make some slight alterations and voila, we have k = F/X.
An entirly valid equation. Not forgetting to change cm to meters we continue.
And like the Glade Plugin's, we plug it in plug it in. So 320 N /
.221 m = k and k = 14.5 N/m
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of Contents
2.
A very stiff spring has a k = 24350 N/m. What is its restoring
force when you distort it 1.00 cm?
(234.5 N)
Here is what you start with:
-
X = 1.00 cm
-
F = ?
-
k = 24350 N/m
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m = ?
-
g = 9.8 m/s/s
-
T = ?
This is again a matter of plugging, hehehe. the formula k = F/x works
great. 24350 N/m = F / (.01 m). Boy, this is tough! multiply
24350 by .01 to get kx. WAIT A MINUTE, we shoould have just used F = kx.
At any rate, the answer comes out the same -- 234.5 N
Go to: ProblemFormulasTable
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3.
Another totally different spring has a spring constant of 249 N/m.
How far will it distort when you hang a 5.400 Kg mass from it?
(.213 m)
Here is what you start with:
-
X = ?
-
F = ?
-
k = 249 N/m
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m = 5.400 Kg
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g = 9.8 m/s/s
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T = ?
Hmmm, I tink that a formula that solves for x would be a good idea!
F/k = x is GRRRRReat! But we must remember that we don't have F!
=(
Fortunately we're brilliant and know that F = mg so mg / k = x.
Put the numbers in and get (5.4 Kg)(9.8 m/s/s)/(249 N/m) = x and therefore
x = .213 m!
Go to: Problem FormulasTable
of Contents
4.
A spring has a constant of 34 N/m, and a 3.00 kg mass hangs from it.
What would be the period of motion of the mass on the spring?
(21.2 seconds)
Here is what you start with:
-
X = ?
-
F = ?
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k = 34 N/m?
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m = 3.00 Kg?
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g = 9.8 m/s/s
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T = ?
Well, we can use the formula F=kx on this one *sniff, sniff*, but
we can use the equation for period, T = 2p(square
root of)k/m!!!!! hmm, what do we do with numbers and an equation?
ah yes, plug them in. so T = 2p(square
root of)(34 N/m)/(3.00 Kg) which yields roughly 21.2 seconds.
Go to: ProblemFormulasTable
of Contents
5.
You need a period of exactly 1.00 seconds, and you have a mass of
230 grams. What must be the spring constant of the spring?
(.0058 N/m)
Here is what you start with:
-
X = ?
-
F = ?
-
k = ?
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m = 230 g
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g = 9.8 m/s/s
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T = 1.00 seconds
I bet the period formula would work splendidly. T = 2p(square
root of)k/m with things plugged is (1.00 sec) = 2p(square
root of)k/(.23 kg). Did you remember to convert the grams to kg...............good.
So work it out, work it out y'all! and you get .0058 N/m, right!?
Go to: ProblemFormulasTable
of Contents
6.
You need a period of exactly 1.00 seconds, and you have a spring constant
of 120. N/m. What must be the mass on the spring (in grams)? (4700000
grams)
Here is what you start with:
-
X = ?
-
F = ?
-
k = 120 N/m
-
m = ?
-
g = 9.8 m/s/s
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T = 1.00 seconds
Hmmm, I would like to venture a guess, that we should use the period formula.
T = 2p(square root of)k/m plug in what we know.....(1.00
sec) = 2p(square root of)(120)/m. Seems
simple enough. Doodamaff (do the math) and you get the magically
delicous number 4700 kg, but in grams that is 4700000 g.
Go to: Problem FormulasTable
of Contents
7.
Joe takes an unknown spring and hangs a 5.000 Kg mass on
it. The weight of the mass stretches the spring 230.3 cm. What
would you expect the period of the mass to be if it were set in motion?
(41
seconds)
Here is what you start with:
-
X = 230.3
-
F = ?
-
k = ?
-
m = 5.000 kg
-
g = 9.8 m/s/s
-
T = ?
I would expect the period to be oh ummm I don't know maybe 41 seconds.
But that doesn't explain why! This problem requires a more complex
approach, but does allow us to use that great formula from the beginning.
We need to find k so we can use it in T = 2p(square
root of)k/m, but in order to do that with what we've got, we must use k
= mg / x. Therefore (5.000 kg)(9.8 m/s/s) / (.2303 m) = 212.765 N/m.
We can now solve the period formula T = 2p(square
root of)k/m. With the stuff plugged it looks like this: T =
2p(square root of)(212.765 N/m)/(5.000 kg)
so T = umm well 41 seconds. It appears that my Jedi senses have yet
again served me well!
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of Contents
And May the Force be with you.................................................................................................................................................................................Always!