Damped Harmonic Motion-Report 1 PDF

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Lab report on Damped Harmonic Motion...

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Lab report: Damped Harmonic Motion Name:

Date: Course:

Question 1: Does the period increase or decrease for a damped system ( b ≠ 0 ) compared to an undamped system ( b=0 )? Explain your reasoning. Ans: The relation between the angular frequency

ω

and time-period T

is:

ωT =2 π Which can be rewritten as:

T= Expression of

√

ω=

2π … . ( i) ω

ω is:

( )

k b − m 2m

2

Hence increasing the damping force that is for higher values of the damping coefficient become smaller. Since ω appears in the denominator in ( i ) and for a damped system and time, the period of the damped system will consequently increase with time.

b , ω will

ω decreases with

Question 2: (optional) Does the decay time increase or decrease as the damping is increased? State your reasoning. Ans: From equation (4), we notice the expression for decay time, τ , contains the damping coefficient b and also carries a negative exponent (-1). Hence the damping coefficient appears in the denominator of the expression of τ . Hence, increasing the damping coefficient will reduce the decay time. This means the system will return to equilibrium faster. This is obvious as damping is a force that resists the motion of the system and increasing it will impede more the motion of the system thus decaying its motion faster that is in a shorter decay time. Question 3: Show that the decay time has the units of time. Ans: The expression of decay time is:

(

τ= First determine the unit of

κ .

b +κ 2m

)

−1

… .. ( i )

κ is:

The expression of

√(

κ=

)

2

b K − 2m m

Unit of

b is kg / s and that of K

The unit of

So, the unit of

is

N /m

m is kg . So, the unit of

or kg / s 2 . Hence the unit of

K m

( ) b 2m

2

is s−2 .

is s−2 .

k is s−1 .

The units of different physical quantities in the above expression are expressed below:

[ b ]= kg / s [ m] =kg

[ k ]= s−1 So, from

( i ) , the unit of

Hence the unit of

b 2m

is

s

−1

. The unit if

k

s τ will be (¿¿−1)−1 or s (− 1 ) ×( −1) ¿

Hence proved that the unit of τ

is also s−1 .

or s .

is seconds that is the unit of time.

Question 4: Show that for the critically damped case, both ω and Ans: The expression for

√(

κ=

)

K

is:

2

b K − 2m m

For critically damped system, we have get:

2 b =4 Km . Substituting this value in the above equation, we

k=

√

¿

4 Km K − 2 m 4m

√

K K − m m ¿0

Since

κ vanish.

ω2 =−k 2 , for k =0 ,

Hence proved.

ω is also 0.

Data 1: Table 1: spring constant

Spring constant dial 1 3 5 7 10

x 0 (m)

m(kg)

k (N /m)

0.28 0.18 0.10 0.08 0.04

0.100 0.100 0.100 0.100 0.100

3.50 5.45 9.81 12.26 24.53

Data 2: Table 2: underdamped harmonic motion

Damping dial

1

k (N /m)

24.53

m (kg)

0.100

t (s)

x max (m)

0 0.58 1.15 1.72 2.27 2.81

0.30 0.24 0.20 0.15 0.11 0.09

Analysis 1: Make a plot of ln x max versus t . From this plot extract the value of the damping coefficient b . Include the graph with your lab report. Ans: The plot is mentioned below:

Equation obtained from the plot is:

X max =0.3109 e−0.439 t … … ( i )

Comparing

( i ) with the standard form, we get: b =0.439 2m b=2× 0.100 ×0.439

b=0.0878 Hence the value of damping coefficient



b=¿ 0.0878

b is 0.0878

kg / s .

kg / s .

Data 3 and Analysis 2: Table 3: frequency for underdamped motion

ωth (rad /s)

15.66

T ¿ (s) 0.58 0.57 0.57 0.55 0.54

ω¿ (rad / s) 10.83 11.02 11.02 11.42 11.64

Data 4: Table 4: overdamped motion

k (N /m)

Damping dial

m (kg)

3.50

At the Maximum Value

0.050

t (s)

x (m)

0 0.08 0.13 0.18 0.24 0.31

0.60 0.50 0.40 0.30 0.20 0.10

Analysis 3: Make a plot of ln x versus t . From this plot extract the decay time τ t1 /2 . Include the graph with your lab report. Ans: The plot is shown below:

The relation found between

ln ( x )

and t

by the plot is:

ln ( x ) =−5.7565 t−0.3042 Comparing with the theoretical equation, we get:

1 =5.7565 τ ¿ , τ=

1 5.7565

¿ , τ=0.17 Hence the value of

τ

is 0.17 s .

and half-life

Relation between

t 1 / 2 and τ

is:

t1 / 2=τ ln ( 2) ¿ , t1 / 2=0.12 Hence the value of

t 1 / 2 is 0.12 s .



τ =¿ 0.17 s



t 1 /2 =¿ 0.12 s.

Question 5: Look up typical resonance frequencies for a human eardrum, a human chest and a human skull. Frequencies are usually given in Hertz (Hz) with 1 Hz equal to 1 cycle per second. Give your answer in both Hz and rad / s . To convert use the fact that one cycle is 2 π radians. Ans: 

Human Eardrum Frequency: 20 Hz (minimum),

20 ×10

3

Hz (Maximum)

2 π ×20=125.66 rad / s (minimum) and 2 π ×20 ×10 =125663.71 rad / s (maximum)

The angular frequency is: 3



Human Chest Frequency: 25 Hz (Male), 33 Hz (Female)

The angular frequency is: 2 π ×25=157.08 rad / s (Female)  

(Male),

2 π ×33=207.35 rad / s

Human Skull Frequency: 972 Hz (minimum), 1230 Hz (maximum) The angular frequency is: 2 π × 972=6107.26 rad / s (minimum), 2 π ×1230 =7728.32 rad / s (maximum)...