Report 5 Forced Oscillation

GENERAL EXPERIMENT REPORT EXPERIMENT 5 : Forced Oscillation NAME STUDENT NUMBER DATE CLASS GROUP CO-WORKER TEACHING ASSISTANT

NGUYEN QUANG HANH 20070933 2007.04.26 & 2007.05.03 THURSDAY 15:30 5 ABUBAKR RASHEED YONGSOP HWANG

Introduction Oscillation is a popular physics phenomenon that has many applications in the society. In the daily life we often see damped oscillations and forced oscillations, and in this experiment, we will get more acquainted with these kinds of oscillations: 1. Simple harmonic oscillation : Oscillation in the nearly ideal condition which has no or negligible resistance. 2. Damped harmonic oscillation: Oscillation added a damping force proportional to the velocity. 3. Forced harmonic oscillation: Oscillation added a sinusoidally varying driving force.

Materials and Methods: 1. Materials:

Other equipments : a ruler, masses

Methods: 1) natural frequency - hang a weight on the spring, and stretch - observe the amplitude and the period shown on the machine. 2) damping oscillation _ hang a weight and also set the magnet to make a friction _ observe the displacement or period 3) resonance _ set the driving frequency and observe the amplitude _ change the driving frequency and find out the resonance frequency 4) transient effect & beat _ set the driving frequency lower or higher than the natural frequency by 0.1Hz _ observe the displacement and the period within the short time. Take note 30 consecutive values of amplitude.

Results (1) Simple Harmonic Oscillation

Mass (g)

Length (mm)

Mass of bar (g)

51.0679

50

51.649

(2) Period (s) 0.42

As the spring is in equilibrium, the net force exerted on it equals to zero : ∑F = mg – kx = 0 k=

= 20.13 (m/s2)

=

ϖ=

=

= 14.0 (rad/s)

The motion is described by the equation : x = A cos (14.0 t + φ).

T=

= 2π

= 2π

=2π

Error :

= 0.449 (s)

100% = 6.46%

(3) Damped Oscillation The damped oscillation is presented by the equation : x=A

cos(ϖt + φ)

With the collected data we can calculate the average value of γ We have the ratio of the two consecutive values of x equals to the difference of time equals to a period T, hence Δt = T.

. The values are collected with

x2/x1 = exp( γ1T) , x3/x2 = exp( γ2T), … xn/x n-1=exp( γn-1T)  xn/x1 = exp [ (γ1 + γ2 + …. + γn-1)T] = exp [

(n 1)γaverageT]

 γaverage = (γ1 + γ2 + …. + γn-1) /(n-1) No magnet :

1 2 3 4 5 6 7 8 9 10

Amplitude (mm)

-γ (s )

120 86 77 63 59 53 44 40 35 30 Average -γ (s )

-0.77 -0.26 -0.47 -0.15 -0.25 -0.43 -0.22 -0.31 -0.36 -0.40

Figure 1 – No magnet

Weak magnet:

1 2 3 4 5 6 7 8 9 10

Amplitude (mm)

-γ (s )

136 74 71 56 49 45 41 35 32 29 Average -γ (s )

-1.42 -0.10 -0.55 -0.31 -0.20 -0.22 -0.37 -0.21 -0.23 -0.45

Figure 2 – Weak Magnet

Strong magnet:

1 2 3 4 5 6 7 8 9 10

Amplitude (mm)

-γ (s )

176 87 76 58 51 42 41 34 32 26 Average -γ (s )

-1.64 -0.31 -0.63 -0.30 -0.45 -0.06 -0.44 -0.14 -0.48 -0.56

Figure 3 – Strong Magnet

(4) Forced Oscillation – Resonance

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Frequency (Hz) 0.70 0.90 1.10 1.30 1.50 1.70 1.90 2.05 2.10 2.15 2.20 2.25 2.30 2.35 2.50

Amplitude (mm) 2 3 3 4 5 8 14 28 45 99 89 41 29 22 10

g

The maximum amplitude recorded in the experiment is at the frequency f= 2.15Hz. The system resonance when the driving frequency and the natural frequency have the same value.

With the calculated and measured values of T in the first experiment, we can calculate the natural frequency of the oscillation according to the equation : f = Calculated value : T= 0.449(s) , f=1/0.449s = 2.23 (Hz) Error :

100% = 3.59%

Measured value : T= 0.42(s) , f=1/0.42s = 2.38 (Hz) Error :

100% = 9.67%

(5) Forced Oscillation – Beat

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Amplitude (mm) 134 76 74 65 64 68 59 56 51 28 23 20 19 22 26

16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Amplitude (mm) 27 29 33 31 34 33 32 32 31 30 29 30 31 30 31

Period(s) 0.43 Resonance Frequency(Hz) 2.15 DrivingFrequency(Hz) 2.05

fbeat= fresonance - fdriving = 2.15 Tbeat = 1/fbeat = 10 (s) The time interval between the two first maxima of the collected amplitude is T1=(18 5.16 s. (The maximum values are recorded at the 6th and 18th orders) The time interval between the two first maxima of the collected amplitude is T2=(13 .44 s. (The minimum values are recorded at the 5th and 13th orders) Taverage=(T1+T2)/2 = 4.3 s Error = (4.3-10)/4.3 = 132.6 % Discussion (1) Damped Oscillation The oscillation is affected by the air resistance in the first case and both air resistance and magnetic force in the second and third cases. The force on the mass connected to the spring is proportional with the velocity v=dx/dt, and equals to –b , where b is a constant that describes the strength of the damping force. The negative sign shows that the net resistance force is always opposite in direction to the velocity. Apply Newton’s second law for the system:

F = -kx - b

=m

(*)

If the damping force is relatively small, the root of the above equation has the form of:

=

A

=

x=A

cos(ϖt + φ)

ϖA

sin(ϖt + φ) =

cos(ϖt + φ)

A

cos(ϖt + φ)

=(

2γϖA ϖ²)x

sin(ϖt + φ)

2γϖA

ϖA

x

ϖ²A

sin(ϖt + φ)

(ϖt + φ)

sin(ϖt + φ)

Besides, from the equation (*), we have:

=  (  γ=

ϖ²) =

ϖA

x and 2γϖ =

sin(ϖt + φ))

ϖ

and ϖ=

 The equation of x: x= A

cos(

The angular frequency ϖ is always smaller than

t + φ)

and becomes zero when b=2

, at which

condition called the critical damping. The system no longer oscillates but returns to the equilibrium position with the function of x= A

cosφ.

When b>2 the condition is called overdamping. The system also returns to the equilibrium position without oscillation but more slowly than in the critical damping. The solution of x in this condition is x = C1 + C2 , where C1 and C2 are constants depending on the initial conditions and α and β are constants identified by m,k and b. When b<2 , the condition is called underdamping and the system oscillates with the decreasing amplitude. All the cases of damping forces in this experiment lead to the underdamping condition. (2) Forced Oscillation and Resonance

Apply a periodically varying driving force with angular frequency ϖ to the damped harmonic oscillation to make a driven oscillation. The applied force is determined according to the equation : F = F0 cos ϖt. The net force is : F=-kx - b Newton’s second law gives : m

=-kx - b

+ F0 cos ϖt.

+ F0 cos ϖt or

=

x

+ cos ϖt (**)

The root of this equation has the form of : x= A = iϖA

,

=

Substitute into the equation (**) : ( A The amplitude of the driven oscillation :

A=

From the data recorded we can see that the amplitude of the driven oscillation in the experiment is about 30 mm. (3) Beat: The time interval between two maxima or two minima of amplitude is Tbeat. In the time period of equal to T beat, the two oscillation perform the numbers of cycles: n and n-1. We have: nTa=Tbeat & (n-1)Tb = Tbeat.  Tbeat = ( Ta Tb ) / (Tb – Ta)  fbeat = (Tb – Ta) / ( Ta Tb ) = 1/Ta – 1/Tb = fa - fb Hence, we can use this formula to calculate fbeat . The data recorded has the values which are not very different from each other, therefore there may be error in determining the actual maxima and minima, which leads to such a big difference between the experimental and theoretical results.

Conclusion : The tool used to measure frequency, amplitude and period may have high accuracy, but most of the errors come during the experiment like measuring the lengths, performing oscillations and recording the data.

References University Physics with Modern Physics 11th Edition (Young & Freedman) 2007 Spring - General Physics Experiment I (KAIST) http://phylab.kaist.ac.kr