What are the Periods of Oscillation for Different Pendulum Configurations?

[JJBrian]

Homework Statement

1. A 0.5 kg mass extends a spring 1 cm. What is the frequency of oscillation of this mass
and spring?
2. A 1 m stick is used as a simple pendulum with a 3 kg weight on the end. What is its
period of oscillation?
3. The same meter stick is used as a rigid pendulum with no weight. What is its period?

Homework Equations

F = −kx = ma.
x = Asin(wt) .
w = sqrt(k/m)
w = 2pie/T
v = Awcos(wt) .
vmax= Aw.
a = −A^2sin(wt) .
amax= wvmax=Aw^2.
t = Ia
t = −FLsin(theta) = −mgLsin(theta) = −mgL(theta)
I = mL^2
theta = Asin((2pie/T)*t)
T = 2pie*sqrt(L/g)
2pie/T =sqrt(3g/2L)

The Attempt at a Solution

1. $w = \sqrt{k/m}$
$w = \sqrt{.01m/.5kg}$
$w = 0.1414~rad/s$

2. $T = 2\pi~\sqrt{L/g}$
$T = 2\pi~\sqrt{1m/9.8m/s^2}$
$T = 2.007~s$

3. $2\pi/T = \sqrt{3g/2L}$
$2\pi/T = \sqrt{3(9.8)/2(1)}$
$T = 2\pi/3.834$
$t = 1.6388~s$

Can someone check my work?

 

[malawi_glenn]

For 1) you need to find the spring constant k which is kx = mg => k = mg/x

 

[kuruman]

You write
T = 2pie*sqrt(L/g)
and then you write
2pie/T =sqrt(3g/2L)

Where did the "3" in the second line come from?

 

[malawi_glenn]

You write
T = 2pie*sqrt(L/g)
and then you write
2pie/T =sqrt(3g/2L)

Where did the "3" in the second line come from?

I think it is because this is problem 3 with a rod, moment of inertia mL²/3

 

[kuruman]

Yes, that's what it is, nevertheless the second line does not follow from the one above it. Farther up, OP says I = mL^2. There is no factor of 3 anywhere that I can see except in the last line. This is sloppy work that needs to be pointed out.

 

[malawi_glenn]

Yes, that's what it is, nevertheless the second line does not follow from the one above it. Farther up, OP says I = mL^2. There is no factor of 3 anywhere that I can see except in the last line. This is sloppy work that needs to be pointed out.

I think that was just for problem 2 where the rod is considered massless.

Oh well, time for sum pie