SMES1201: VIBRATION AND WAVES (GROUP 8)

TUTORIAL #2

NAME : WAN MOHAMAD FARHAN BIN AB RAHMAN MATRICES NO : SEU110024 DATE : 12 MAY 2014 QUESTION 1

A damped oscillator with mass 2 kg has the equation of motion 2x + 12 x + 50 x = 0 , where x is the displacement from equilibrium, measured in meters.

(a) What are the damping constant and the natural angular frequency for this oscillator? (b) What type of damping is this ? Is the motion still oscillatory and periodic ? If so, what

is the oscillation period ?

Solution: (a) Mass of oscillator = 2 kg Equation of motion for damped oscillation :

Fnet = Fdamping + Frestore

That is , x + 𝛾

𝑚 x + 𝜔o

2x = 0 ----------------- (*)

Given that the equation of motion is 2x + 12 x + 50 x = 0 --------------------------(**) We then divide (**) by 2, become x + 6 x + 25 x = 0 --------------------------(***) By comparison of (*) and (***) , 𝛾

𝑚 = 6s-1 , then damping constant, 𝛾 = 6s-1 (2kg) = 12kgs-1 ,

𝜔o

2 = 25 , then natural angular frequency , 𝜔o = 5 s-1

(b) We stipulate solution x(t) to equation (*) for a damped oscillator of the form 𝑒𝛼𝑡

times X(t)

x (t) = 𝑒𝛼𝑡 X (t)

x (t) = 𝛼𝑒𝛼𝑡 X (t) + 𝑒𝛼𝑡 X (t)

x (t) = 𝛼2𝑒𝛼𝑡 X (t) + 2𝛼𝑒𝛼𝑡 X (t) + 𝑒𝛼𝑡X (t)

From the types of damping, we test for value of 2m𝜔o that is equal to (2)(2kg)(5s-1) = 20kgs-1 , which is larger than 𝛾 = 12kgs-1 , thus the oscillation described an underdamped oscillation.

The oscillation occur and periodic such that the amplitude decreases gradually by an exponential factor of 𝑒−5𝑡 .

The oscialltion period, T = 2𝜋

𝜔

For underdamped oscillator , x(t) = Ao 𝑒−𝛾𝑡/2𝑚 sin (𝜔𝑡 + 𝜑o ) ,

Where 𝜔 = ( 𝜔o 2 - 𝛾2 / 4m2 ) ½ = ( 𝜔o

2 - 1

4 ( 𝛾 / m ) 2 ) ½ = ( 25 -

1

4 ( 6 ) 2 ) ½ = 4s-1 ( > 0 )

Thus, T = 2 𝜋 / 4s-1 = 1.5708 s

QUESTION 2

A block of mass 90 grams attached to a horizontal spring is subject to a friction force F fric = -γx when in motion , with γ = 0.18 kg s-1 . The system oscillates about its equilibrium position but loses 95% of its total mechanical energy during one complete cycle.

(a) Determine the period of the damped oscillation (b) Find the natural angular frequency and the spring constant , k

Solution (a) Mass of block = 90 grams = 0.09 kg

F fric = -γx , with 𝛾 = 0.18 kgs-1

The block undergoes underdamped oscillation and the system maintains 5% of its total mechanical energy.

Time dependent amplitude, A(t),

A(t) = Ao 𝑒−𝛾𝑡/2𝑚 = Ao 𝑒

−𝑡

2𝑚/𝛾 = Ao 𝑒−

𝑡

2(0.09𝑘𝑔)/(0.18𝑘𝑔𝑠−1) = Ao 𝑒−𝑡 = Ao (

1

𝑒𝑡 )

Etotal ∝ A(t) 2 , initially Eo ∝ Ao2 , so Etotal ∝ ( Ao 𝑒−𝑡

) 2 = Ao2

𝑒−2𝑡 ===> Etotal / Eo = 𝑒−2𝑡

Solving this for the time when Etotal / Eo = 0.05 By comparison , 0.05 = 𝑒−2𝑡

, thus t = 1.498 s

The system oscillates about its equilibrium position but loses 95% of its total mechanical energy during ONE COMPLETE CYCLE, means that : The period of the damped oscillation, T = t = 1.498 s

(b) The period of the damped oscillations, T = 2𝜋

𝜔 , thus 𝜔 =

2𝜋

T

Angular frequency, 𝜔 = 2𝜋

1.498s = 4.194 s-1

To find the natural angular frequency, 𝝎o , use the formula of 𝜔 = ( 𝜔o 2 -

1

4 ( 𝛾 / m ) 2 ) ½

Thus, , 𝜔o = ( 𝜔 2 +

1

4 ( 𝛾 / m ) 2 ) ½ = ( (4.194)

2 + 1

4 ( 0.18 / 0.09 ) 2 ) ½

𝜔o = 4.312 s-1 Spring constant , k = 𝜔o

2 . m = ( 4.312) 2 ( 0.09 ) = 1.673 Nm-1

QUESTION 3

An oscillator with a mass of 300 g and a natural angular frequency 0f 10 s-1 is damped by a force -24 x N, and driven by a harmonic external force with amplitude 1.2 N and angular frequency 6s-1 .

(a) Write the equation of motion for this system. (b) Calculate the amplitude A and phase lag δ of the steady-state displacement ,

Xp (t) = A cos (𝜔t – δ ).

Solution: (a) External or driving force : F(t) = Fo cos (𝜔e t)

Equation of motion for forced oscillations: Fnet = Fdamping + Frestore + F(t)

That is , x + 𝛾

𝑚 x + 𝜔o

2x = Fo

𝑚 cos (𝜔e t) ----------------- (*)

Given mass of oscillator ,m = 300g = 0.3 kg 𝜔o = 10 s-1

Fdamping = -24 x N, with 𝛾 = 24kgs-1

Fo = 1.2 N , 𝜔e = 6s-1

Then, equation of (*) will become : , x + 24

0.3 x + 10

2x = 1.2

0.3 cos (6 t)

We simplify it, become x + 80 x + 100 x = 4 cos (6 t)

(b) We hypothesize that :

Xp (t) = A cos (𝜔et – δ ) xp (t) = - 𝜔e A sin (𝜔et – δ ) xp (t) = - 𝜔e 2A cos (𝜔et – δ )

We know that , the displacement amplitude of a driven oscillator in the steady state,

A = Fo

𝑚 / ( (𝜔o

2 - 𝜔e2 ) 2 + 𝛾2 𝜔e

2 / m2 ) ½

Also the phase angle or phase lag between force and displacement in the steady state,

tan δ = ( 𝛾 𝜔e / m ) / ( 𝜔o

2 - 𝜔e2 )

Then, A = 1.2 N

0.3 𝑘𝑔 / ( (10

2 - 62 ) 2 + 242 62 / 0.32 ) ½

= 4 / √234496 = 0.00826 m Phase lag (in radians) tan δ = ( 24 (6) / 0.3 ) / ( 10

2 - 62 ) tan δ = 480 / 64 = 7.5 δ = 1.438 radians

QUESTION 4

A block of mass m = 200 g attached to a spring with constant k = 0.85 N m-1 . When in motion, the system is damped by a force that is linear in velocity, with damping constant, γ = 0.2kg s-1 . Then, this block+ system system is subjected to a harmonic external force F(t) = Fo cos (𝜔e t) , with a fixed amplitude Fo = 1.96 N.

(a) Calculate the driving frequency 𝜔e, max at which amplitude resonance occurs. Find the steady-state displacement amplitude.

(b) Calculate the steady-state displacement of the system when the driving force is constant, i.e. for 𝜔e = 0

Solution : (a) Mass of block , m = 200 g = 0.2 kg

Spring constant, k = 0.85 Nm-1

Damping constant, 𝛾 = 0.2kgs-1

Harmonic external force , F(t) = Fo cos (𝜔e t) , given Fo = 1.96 N dA

dωe = 0 (consider Fo , 𝛾 , m, mo = constant)

A = function of 𝜔e

U (𝜔e) = (𝜔o 2 - 𝜔e

2 ) 2 + 𝛾2 𝜔e2 / m2

A = Fo

𝑚 U-1/2

dA

dωe =

dA

dU x

dU

dωe = +𝜔e

Fo

𝑚 U-3/2 ( 2(𝜔o

2 - 𝜔e2 ) 2 - 𝛾2 /𝑚2 )

Thus, max A ---------> 2(𝜔o 2 - 𝜔e

2 ) 2 = 𝛾2 /𝑚2

𝜔e = (𝜔o 2 - 𝛾2 /2𝑚2 ) ½

= (3.75)1/2

= 1.936 s-1

𝜔o = (k/m)1/2

A = Fo

𝑚 / ( (𝜔o

2 - 𝜔e2 ) 2 + 𝛾2 𝜔e

2 / m2 ) ½

= (1.96/0.2) / ( (2.06155 2 - 1.9362 ) 2 + (0.2)2 (1.9362 / (0.2)2 ) ½

= 4.9 m

𝜔e, max < 𝜔o , that means amplitude resonance always occurs for a

driving frequency < 𝜔o

(b) When the driving force is constant ( 𝜔e = 0 ), then this is the case when the external

force varies much more slowly than natural restoring force , Frestore = -m 𝜔o 2 x.

Thus, 𝜔e can be ignored relative to 𝜔o .

From formula of tan 𝛿 = ( 𝛾 𝜔e / m ) / ( 𝜔o

2 - 𝜔e2 ) , substitute 𝜔e = 0

It will become tan 𝛿 = ( 𝛾 𝜔e

/ m ) / ( 𝜔o 2 ) , and since 𝜔e

/ 𝜔o 2 is very small number,

Then tan 𝛿 ≈ 0

From formula of A = Fo

𝑚 / ( (𝜔o

2 - 𝜔e2 ) 2 + 𝛾2 𝜔e

2 / m2 ) ½ , substitute 𝜔e = 0

It will become A = Fo

𝑚 / ( (𝜔o

4 + 0 ) ½ = Fo

𝑚 / 𝜔o

2 = Fo / m𝜔o 2

The steady-state displacement of the system, = ( Fo / m𝜔o 2 ) cos (𝜔e t)

= ( Fo / m𝜔o 2 ) (1)

= 1.96 / 0.2(2.06155)2 = 2.306 m