MIT Chapter 5 Solutions

Vibrations and Waves MP205, Assignment 5 Solutions 1. Solve the steady-state motion of a forced oscillator (with no resistive force) if the driving force is of the form F = F0 sin(ωt). For a forced oscillator with driving force F0 sin(ωt), it’s equation of motion is d2 x + kx = F0 sin(ωt) dt2 d2 x F0 + ω02 x = sin(ωt) 2 dt m

m

Using the complex exponential method, let   x = A cos(ωt − δ) = Re Aei(ωt−δ) = Re(z) z = Aei(ωt−δ) dz = iAωei(ωt−δ) dt d2 z = −Aω 2 ei(ωt−δ) dt2    π π = Re ei(ωt− 2 ) sin(ωt) = cos ωt − 2 So we can write our equation as: F0 i(ωt− π ) d2 z 2 + ω02 z = e 2 dt m F0 i(ωt− π ) 2 −Aω 2 ei(ωt−δ) + ω02 ei(ωt−δ) = e m F0 iωt −i π −Aω 2 eiωt e−iδ + ω02 eiωt e−iδ = e e 2 m F0 −i π −Aω 2 e−iδ + ω02 e−iδ = e 2 m F0 −i π iδ −Aω 2 + ω02 A = e 2e m F0 i(− π +δ) e 2 A(ω02 − ω 2 ) = m  π   π  F0 F0 = cos − + δ + i sin − + δ m 2 m 2 Comparing real and imaginary coefficients:  π  F0 cos − + δ m 2   F0 π 0= sin − + δ m 2  π  sin − + δ = 0 2 A(ω02 − ω 2 ) =

F0 m

 π  cos − + δ 2 ω02 − ω 2  π  ⇒ sin − + δ = 0  2π 2 ⇒δ= 3π ⇒A=

2

A= = =

F0 m

 π π cos − + 2 2 ω02 − ω 2 F0 m

ω02 − ω 2

cos (0)

F0 m

ω02 − ω 2 F0 m

  π 3π cos − + A= 2 2 2 ω0 − ω 2 =

F0 m

ω02 − ω 2

=− To ensure A is positive then for δ =

π 2

cos (π)

F0 m

ω02 − ω 2

we require ω0 > ω, and for δ =

3π 2

we require ω0 < ω.

x = A cos(ωt − δ) where: F0 A = 2 m 2 ω0 − ω  π when ω0 > ω 2 δ= 3π when ω0 < ω 2 2. An object of mass 0.2 kg is hung from a spring whose spring constant is 80 Nm−1 . The body is subject to a resistive force given by −bv, where v is its velocity in ms−1 and b = 4 Nm−1 sec. (a) Set up the differential equation of motion for free oscillations of the system, and find the period of such oscillations. (b) The object is subjected to a sinusoidal driving force given by F (t) = F0 sin(ωt), where F0 = 2 N and ω = 30 sec−1 . In the steady state, what is the amplitude of the forced oscillation? (c) What is the mean power input? (d) Show that the energy dissipated against the resistive force in one cycle is 0.063J (a) We know the mass is subject to a resistive force −bv, as well as a force due to the spring−kx: F = −kx − bv ma + bv + kx = 0 0.2a + 4v + 80x = 0 a + 20v + 400x = 0 d2 x dt2

+ 20

dx + 400x = 0 dt

Comparing this to the general form: d2 x dx +γ + ω02 x = 0 2 dt dt We can read off values for γ and ω0 : γ = 20 ω02 = 400

ω0 = 20

To obtain the period of oscillation we require ω: γ2 400 ω 2 = ω02 − = 400 − = 400 − 100 = 3 4 √ 4 √ ω = 300 = 10 3 2π 2π π T = = √ = √ ≈ 0.36s ω 10 3 5 3 (b) For this system the equation of motion is given by: d2 x dx F0 +γ + ω02 x = sin ωt 2 dt dt m Using the complex exponential method, let   x = A cos(ωt − δ) = Re Aei(ωt−δ) = Re(z) z = Aei(ωt−δ) dz = iAωei(ωt−δ) dt d2 z = −Aω 2 ei(ωt−δ) dt2    π π = Re ei(ωt− 2 ) sin(ωt) = cos ωt − 2 So we can write our equation as: d2 z dz F0 i(ωt− π ) 2 +γ + ω02 z = e 2 dt dt m F0 i(ωt− π ) 2 −Aω 2 ei(ωt−δ) + γiωAei(ωt−δ) + ω02 ei(ωt−δ) = e m F0 iωt −i π −Aω 2 eiωt e−iδ + iγωAeiωt e−iδ + ω02 eiωt e−iδ = e e 2 m F0 −i π −Aω 2 e−iδ + iγωAe−iδ + ω02 e−iδ = Ae 2 m F0 −i π iδ −Aω 2 + iγωA + ω02 A = e 2e m F0 i(− π +δ) A(ω02 − ω 2 ) + iγωA = e 2 m  π   π  F0 F0 = cos − + δ + i sin − + δ m 2 m 2

Comparing real and imaginary coefficients: A(ω02 − ω 2 ) = ⇒A= γωA = ⇒A= A=A⇒

F0 m

 π  F0 cos − + δ m 2 F0  π  m cos − + δ 2 ω02 − ω 2   F0 π sin − + δ m 2  π  F0 sin − + δ mγω 2   F0 π sin − + δ mγω 2  π  1 sin − + δ γω 2

 π  cos − + δ = 2 ω02 − ω 2  π  1 cos − + δ = 2 ω02 − ω 2
sin − π2 + δ γω
= 2 π ω0 − ω 2 cos − 2 + δ  π  γω tan − + δ = 2 2 ω0 − ω 2

let θ = − π2 + δ

If we draw this angle on a right angles triangle and use Pythagoras’ theorem:

This gives us:  π  ω2 − ω2 cos − + δ = cos θ = p 2 0 2 (ω0 − ω 2 )2 + (γω)2 Using this in our expression for A: F0  π  A = 2 m 2 cos − + δ 2 ω0 − ω =

F0 m

ω02 − ω 2 p ω02 − ω 2 (ω02 − ω 2 )2 + (γω)2 F0

m =p 2 (ω0 − ω 2 )2 + (γω)2

The amplitude is given by: F0 m A= p 2 (ω0 − ω 2 )2 + (γω)2

From (a) we know that: ω0 = 20, m = 0.2 and γ = 20. We’re told here that ω = 30 and F0 = 2, using this in our equation for the amplitude gives: 2 0.2 A= p ((20)2 − (30)2 )2 + ((20)(30))2 10 =p ((400 − 900)2 + (600)2 10 =p ((−500)2 + 360, 000 10 =√ 250, 000 + 360, 000 10 =√ 610, 000 = 0.0128m

(c) The mean power is given by: F 2 ω0 P¯ = 0  2kQ ω0 Using the same values we used in (b), Q =

ω0 γ

1 2

1 Q2

ω

−

ω ω0

=

20 20

= 1 and k = ω02 m = (20)2 (0.2) = 80 we get:

(2)2 (20) P¯ = 2(80)(1)

+

1 20 30

−


30 2 20

+

1 1

4(20) 1
160 − 5 2 + 1 6 1 80 = 160 25 36 + 1 1 1 = 61 2 36 1 36 = 2 61 18 = 61 ≈ 0.3W =

(d) The energy lost per cycle is given by: E = P¯ T = P¯

2π 2π = (0.3) = 0.063J ω 30

* 3. A block of mass m is connected to a spring, the other end of which is fixed. There is also a viscous damping mechanism. The following observations have been made on the system: (1) If the block is pushed horizontally with a force equal to mg, the static compression of the spring is equal to h. (2) The viscous resistive force is equal to mg if the block moves with a certain known speed u. (a) For this complete system (including both spring and damper) write the differential equation governing horizontal oscillations of the mass in terms of m,g, h and u.

(a) (1) tells us: kx x=h = mg kh = mg mg k= h (2) tells us: bv v=u = mg bu = mg mg b= u Using this in: ma + bv + kx = 0 m

d2 x dt2

+b

dx + kx = 0 dt

gives: m

d2 x mg dx mg + + x=0 dt2 u dt h d2 x g dx g + + x=0 dt2 u dt h

Comparing this with d2 x dx +γ + ω02 x = 0 dt2 dt gives us values for γ and ω0 : g u g ω02 = h √ Answer the following for the case that u = 3 gh γ=

r ⇒ ω0 =

g h

(b) What is the angular frequency of the damped oscillations? q (c) After what time, expressed as a multiple of hg , is the energy down by a factor 1e ? (d) What is the Q of this oscillator? √ Using u = 3 gh gives us: g g 1 γ= = √ = u 3 3 gh g 2 ω0 = h (b) To find the angular frequency we use: ω 2 = ω02 −

r

γ2 4 .

γ2 4 g 1 g = − h 9 4h g g = − h 36h 35g = 36h r 35g ω= 36h

ω 2 = ω02 −

g h

(c) The energy decreases according to E(t) = E0 e−γt E(t) = E0 e−γt we need to find a τ st: E(τ ) = E0 e−1 = E0 e−γτ this tells us 1 = γτ 1 τ= γ s =3

h g

So the time taken for the energy to decrease a factor of

1 e

q is t = 3 hg s.

(d) Q=

ω0 γ q

g h

= q 1 3

=

g h

1 1 3

=3 4. A mass m is subject to a resistive force −bv but no springlike restoring force. * (a) Show that its displacement as a function of time is of the form: b x = C − vγ0 e−γt where γ = m (b) At t = 0 the mass is at rest at x = 0. At this instant a driving force F = F0 cos ωt is switched on. Find the values of A and δ in the steady-state solution x = A cos(ωt − δ) (c) Write down the general solution [The sum of parts (a) and (b)] and find the values of C and v0 from the conditions that x = 0 and dx dt = 0 at t = 0 (a) F = −bv ma = −bv b a = − v = −γv m Now, we know that a =

dv dt :

dv = −γv dt 1 dv = −γdt v Integrating both sides gives Z

Z Z 1 dv = −γdt = −γ dt v ln(v) = −γt + D where D is a constant v = e−γt+D v(t) = eD e−γt

Let v0 be the inital velocity at the time 0: v(0) = eD e0 = v0 ⇒ e D = v0 This gives us a final expression for v v(t) = v0 e−γt To get an expression for x we use the fact that v =

dx dt

dx =v dt dx = v0 e−γt dt dx = v0 e−γt dt Integrating both sides gives Z

Z dx =

v0 e−γt dt = v0

v0 −γt e +C γ v0 ⇒ x = C − e−γt γ x=−

Z

e−γt dt

where C is a constant

(b) ma + bv = F0 cos ωt d2 x

dx = F0 cos ωt dt dx F0 d2 x +γ = cos ωt 2 dt dt m

m

dt2

+b

Looking at the steady state solution: x = A cos(ωt − δ), we want to obtain expressions for A and δ. Using the complex exponential method, let   x = A cos(ωt − δ) = Re Aei(ωt−δ) = Re(z) z = Aei(ωt−δ) dz = iAωei(ωt−δ) dt d2 z = −Aω 2 ei(ωt−δ) dt2
cos (ωt) = Re eiωt So we can write our equation as: d2 z dz F0 iωt +γ = e 2 dt dt m F0 iωt −Aω 2 ei(ωt−δ) + γiAωei(ωt−δ) = e m F0 iωt iωt −Aω 2 eiωt e−iδ + Aiγωeiωt e−iδ = e e m F0 −Aω 2 e−iδ + Aiγωe−iδ = m F0 iδ 2 −Aω + Aiγω = e m F0 F0 −Aω 2 + Aiγω = cos (δ) + i sin (δ) m m

Comparing real and imaginary coefficients: F0 cos (δ) m F0 cos (δ) ⇒A=− mω 2 F0 Aγω = sin (δ) m F0 ⇒A= sin (δ) γωm F0 F0 A=A⇒− cos (δ) = sin (δ) mω 2 γωm 1 1 − 2 cos (δ) = sin (δ) ω γ sin (δ) γω γ =− 2 =− cos (δ) ω ω −Aω 2 =

ω cos(δ) = − p γ 2 + ω2 γ sin(δ) = p γ 2 + ω2 Note: we’ve chosen the signs here to ensure we have a positive value for A. Using this to get an expression for A: F0 cos(δ) mω 2 F0 ω p = mω 2 γ 2 + ω 2 F p 0 = mω γ 2 + ω 2

A=−

(c) The general solution is give by x = C −

v0 −γt γ e

+ A cos(ωt − δ)

We have the initial conditions: x(0) =

dx dt (0)

=0

v0 −γt e + A cos(ωt − δ) γ v0 x(0) = C − e0 + A cos(0 − δ) = 0 γ x(t) = C −

v0 + A cos(δ) = 0 γ dx (t) = v0 e−γt − ωA sin(ωt − δ) dt dx (0) = v0 e0 − ωA sin(0 − δ) = 0 dt v0 + ωA sin(δ) = 0

C−

v0 = −ωA sin(δ) = −ωA

γ

!

p γ 2 + ω2 γω

= −A p γ 2 + ω2 γωA F p 0 p =− mω γ 2 + ω 2 γ 2 + ω 2 F0 γ =− m(γ 2 + ω 2 ) And to find C: C−

v0 + A cos(δ) = 0 γ v0 − A cos(δ) C= γ F0 γ m(γ 2 +ω 2 )

F ω p p 0 2 2 2 γ mω γ + ω γ + ω2 F0 F0 + =− m(γ 2 + ω 2 ) m(γ 2 + ω 2 ) =0 =−

+

* 5. The graph shows the power resonance curve of a certain mechanical system when driven by a force F0 sin(ωt), where F0 =constant and ω is variable.

(a) Find the numerical values of ω0 and Q for this system. (b) The driving force is turned off. After how many cycles of free oscillation is the energy of the system down to 1/e5 of its initial value? (e = 2.718) (To a good approximation, the period of free oscillation can be set equal to 2π/ω0 .)

(a) Here we use the fact that with width of the power-resonance curve at half-height ≈ γ ω0 =γ≈2 Q ω0 40 Q= = = 20 2 2 (b) The energy decreases according to E(t) = E0 e−γt E(t) = E0 e−γt we need to find a τ st: E(τ ) = E0 e−5 = E0 e−γτ this tells us 5 = γτ 5 5 τ = = = 2.5 γ 2 The time taken to complete one cycle is

2π ω0 ,

so the time taken to complete n cycles is n ω2π0 :

τ = 2.5 = n

2π ω0

2π = 2.5 ω0 2π n = 2.5 40 n(0.16) = 2.5 2.5 n= = 15.6 ≈ 16 0.16

⇒n

6. The figure shows the mean power input P¯ as a function of driving frequency for a mass on a spring with damping. (Driving force = F0 sin(ωt), where F0 is held constant and ω is varied.) The Q is high enough so that the mean power input, which is maximum at ω0 , falls to half-maximum at the frequencies 0.98ω0 and 1.02ω0 .

(a) What is the numerical value of Q? (b) If the driving force is removed, the energy decreases according to the equation E = E0 e−γt . What is the value of γ? (c) If the driving force is removed, what fraction of the energy is lost per cycle?

(a) ω0 = γ ≈ width of the power-resonance curve at half-height Q ω0 = 1.02ω0 − 0.98ω0 Q = 0.04ω0 ω0 Q= 0.04ω0 1 = = 25 0.04 (b) From (a) we can just write down the value of γ: γ = 0.04ω0 (c) The energy decreases according to the equation E = E0 e−γt , so the fraction of energy lost is

E E0 :

E0 e−γt E = E0 E0 −γt =e = e−0.04ω0 t The time taken for one cycle is the perios T =

2π ω0 ,

so the fraction of energy lost per cycle is 

E −0.04ω0 =e E0 = e−0.08π s

2π ω0



A new system is made in which the spring constant is doubled, but the mass and the viscous medium are unchanged, and the same driving force F0 sin(ωt) is applied. In terms of the corresponding quantities for the original system, find the values of the following: (d) The new resonant frequency ω00 . (e) The new quality factor Q0 . 0 . (f ) The maximum mean power input P¯m

(g) The total energy of the system at resonance, E00 . (d) For the original system: r

k m

r

k0 = m

ω0 = For the new system: ω00 =

r

r √ 2k √ k = 2 = 2ω0 m m

(e) For the original system: Q=

ω0 γ

Q0 =

√ ω0 √ ω00 = 2 = 2Q γ γ

For the new system:

(f) For the original system: QF02 P¯max = 2mω0 For the new system: 0 P¯max

√ Q0 F02 QF02 2QF02 √ = = = P¯max = 2mω00 2mω0 2m 2ω0

(g) Originally we had E = E0 e−γt , therefore E0 = Eeγt . As there is no k dependence here we see that E00 = Eeγt = E0 .