final take home

8/13/2019 Final Take Home

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[A]

1) The equilibrium displacement for the branch with the eagle on it.

DR GREY: We talked about me doing it this way in calculations because the equilibrium line is

curved on the graph.

Using Hookes law, and the k constant from (1.3) we get the following for the equilibrium

displacement:

2(6 )(9.81m s )1.38 m

( 42.4230906656 N m)

mg k y

kg

=

=

(1.1)

2) The effective k for the branch by applying Hookes law to the motion.

DR GREY: This is how we discussed that I solve this problem.

First we use the result of (1.2), the birds weight of 6 kg, and the following formula:2

Tk

m

= .

( )

2

2 2

2

2

2

2 26kg

2.3629489603s

N42.4230906656

m

Tk

m

k

m T

k

m T

k mT

=

=

=

= =

=

(1.2)

3) The period of the motion.

The period was determined from the graph measuring peak to peak, over the two visible

peaks. They are 2.5 cm apart for both sets of peaks. From 0s to 10s on the graph measures

10.58 cm.

10 s2.5 cm =2.3629489603 s

10.58 cm

(1.3)

Therefore the period is 2.363 seconds.

4) The angular frequency, .

0

2 22.659044022rad s

2.3629489603T s

= = = (1.4)


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5) The damping factor, a, in A=A0e-at

Measuring the amplitude at 0s to be 0.010m and the amplitude at 10s to be 0.15m with a

ruler, we can plug these into the A=A0e-atequation to solve for the damping factor a:

( ) ( ) (10)

1

0.15 m 0.010 m

ln(0.15m / 0.010m)

10s

0.271s

ae

a

a

=

=

=

(1.5)

6) The damping coefficient in the force law F=-bv

We know that a is equal to( )1/ 2

m

b

because of the equation:

( )1/2

0

t

A A e

= , wherem

b= .

With this we can use ato solve for b.

( )0.271

2 6kg

3.252kg/s

b

b

=

=

(1.6)


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[C]

[1] Determine the wavelength for the three frequencies:

25.73 -1

42.875 -1

60.03 -1

343m s13.33 m

25.73 s

343m s8 m

42.875 s

343m s5.71 m

60.03 s

Hz

Hz

Hz

c

f

=

= =

= =

= =

[2] From the pattern, determine if the cave is open or closed at the other end:

This cave is close ended because it follows the following pattern for the closed pipe:1

4L = ,

giving 10m for the length each time:

3(13.34 m) 10.005m

4

5(8 m) 9.9993m

4

7(5.71 m) 10m

4

=

=

=

[3] What is the depth of the cave?

From the results of [2] we see that the cave is 10m long.

[4] Calculate the fundamental resonant frequency and determine the mode numbers for the

measured frequencies (1,2,3 etc.)

Using1

4L = we can get the wavelength for the fundamental frequency and use

c

f = with

c being the speed of sound of 343 m/s to calculate the fundamental frequency.

343m s8.575Hz

4 4(10m)

cf

L= = =

The modes can be seen in [2], going from the highest frequency to lowest they are 3, 5 and 7.


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final take home

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