mit chapter 5 solutions

7/17/2019 MIT Chapter 5 Solutions

http://slidepdf.com/reader/full/mit-chapter-5-solutions 1/13

Vibrations and Waves MP205, Assignment 5 Solutions

1. Solve the steady-state motion of a forced oscillator (with no resistive force) if the drivingforce is of the form F = F 0 sin(ωt).

For a forced oscillator with driving force F 0 sin(ωt), it’s equation of motion is

md2x

dt2 + kx = F 0 sin(ωt)

d2x

dt2 + ω2

0x = F 0m

sin(ωt)

Using the complex exponential method, let

x = A cos(ωt − δ ) = Re

Aei(ωt−δ)

= Re(z)

z = Aei(ωt−δ)

dz

dt = iAωei(ωt−δ)

d2z

dt2 = −Aω2ei(ωt−δ)

sin(ωt) = cos

ωt − π

2

= Re

ei(ωt−

π

2)

So we can write our equation as:

d2z

dt2 + ω2

0z = F 0m

ei(ωt−π

2)

−Aω2ei(ωt−δ) + ω20ei(ωt−δ) =

F 0m

ei(ωt−π

2)

−Aω2eiωte−iδ + ω20eiωte−iδ = F 0

meiωte−i

π

2

−Aω2e−iδ + ω20e−iδ =

F 0m

e−iπ

2

−Aω2 + ω20A =

F 0m

e−iπ

2 eiδ

A(ω20 − ω2) =

F 0m

ei(−π

2+δ)

= F 0m

cos−π

2 + δ

+ i

F 0m

sin−π

2 + δ

Comparing real and imaginary coefficients:

A(ω20 − ω2) =

F 0m

cos−π

2 + δ

⇒ A =

F 0m

ω20 − ω2

cos−π

2 + δ

0 = F 0m

sin−π

2 + δ

⇒ sin

−π

2 + δ

= 0

sin−π

2 + δ

= 0 ⇒ δ =

π23π2



A =F 0m

ω20 − ω2

cos−π

2 +

π

2

=F 0m

ω20 − ω2

cos (0)

=F 0m

ω2

0 − ω2

A =F 0m

ω20 − ω2

cos

−π

2 +

3π

2

=F 0m

ω20 − ω2

cos (π)

= −F 0m

ω20 − ω2

To ensure A is positive then for δ = π2 we require ω0 > ω, and for δ = 3π

2 we require ω0 < ω.

x = A cos(ωt − δ )where:

A =

F 0m

ω20 − ω2

δ =

π2 when ω0 > ω3π2 when ω0 < ω

2. An object of mass 0.2 kg is hung from a spring whose spring constant is 80 Nm−1. Thebody is subject to a resistive force given by −bv, where v is its velocity in ms−1 andb = 4 Nm−1sec.

(a) Set up the differential equation of motion for free oscillations of the system, and findthe period of such oscillations.

(b) The object is subjected to a sinusoidal driving force given by F (t) = F 0 sin(ωt), whereF 0 = 2 N and ω = 30 sec−1. In the steady state, what is the amplitude of the forcedoscillation?

(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

d2x

dt2 + 20

dx

dt + 400x = 0

Comparing this to the general form:

d2x

dt2 + γ

dx

dt + ω2

0x = 0

We can read off values for γ and ω0:

γ = 20ω20 = 400 ω0 = 20



To obtain the period of oscillation we require ω :

ω2 = ω20 −

γ 2

4 = 400− 400

4 = 400− 100 = 3

ω =√

300 = 10√

3

T = 2π

ω

= 2π

10√ 3=

π

5√ 3 ≈0.36s

(b) For this system the equation of motion is given by:

d2x

dt2 + γ

dx

dt + ω2

0x = F 0m

sin ωt

Using the complex exponential method, let


Aei(ωt−δ)

= Re(z)

z = Aei(ωt−δ)

dzdt

= iAωei(ωt−δ)

d2z


sin(ωt) = cos

ωt − π

2

= Re

ei(ωt−

π

2)


d2z

dt2 + γ

dz

dt + ω2

0z = F 0m

ei(ωt−π

2)

−Aω2ei(ωt−δ) + γiωAei(ωt−δ) + ω2

0ei(ωt−δ) = F 0

m

ei(ωt−π

2)

−Aω2eiωte−iδ + iγωAeiωte−iδ + ω20eiωte−iδ =

F 0m

eiωte−iπ

2

−Aω2e−iδ + iγωAe−iδ + ω20e−iδ =

F 0m

Ae−iπ

2

−Aω2 + iγωA + ω20A =

F 0m

e−iπ

2 eiδ

A(ω20 − ω2) + iγωA =

F 0m

ei(−π

2+δ)

= F 0m

cos−π

2 + δ

+ i

F 0m

sin−π

2 + δ




A(ω20 − ω2) =

F 0m

cos−π

2 + δ

⇒ A =F 0m

ω20 − ω2

cos−π

2 + δ

γωA = F 0m

sin−π

2 + δ

⇒ A =

F 0mγω

sin−π

2 + δ

A = A ⇒F 0m

ω20 − ω2

cos−π

2 + δ

=

F 0mγω

sin−π

2 + δ

1

ω20 − ω2

cos−π

2 + δ

=

1

γω sin

−π

2 + δ

sin−π

2 + δ

cos −π2 + δ

= γω

ω20 − ω2

tan−π2 + δ

= γωω2

0 − ω2

let θ = −π2 + δ

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

This gives us:

cos−π

2 + δ

= cos θ =

ω20 − ω2

(ω20 − ω2)2 + (γω)2

Using this in our expression for A:

A =F 0m

ω20 − ω2

cos−π

2 + δ

=F 0m

ω20 − ω2

ω20 − ω2

(ω20 − ω2)2 + (γω)2

=F 0m

(ω20 − ω2)2 + (γω)2

The amplitude is given by:

A =F 0m

(ω20 − 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 F 0 = 2,using this in our equation for the amplitude gives:

A =20.2

((20)2 − (30)2)2 + ((20)(30))2

= 10

((400 − 900)2 + (600)2

= 10

((−500)2 + 360, 000

= 10√

250, 000 + 360, 000

= 10√ 610, 000

= 0.0128m

(c) The mean power is given by:

P̄ = F 20 ω0

2kQ

1ω0ω − ω

ω0

2+ 1

Q2

Using the same values we used in (b), Q = ω0γ

= 2020 = 1 and k = ω2

0m = (20)2(0.2) = 80 we get:

P̄ = (2)2(20)

2(80)(1)

12030 − 30

20

2+ 1

1

= 4(20)

160

1

−56

2+ 1

= 80160

12536 + 1

= 1

2

16136

= 1

2

36

61

= 18

61≈ 0.3W

(d) The energy lost per cycle is given by:

E = P̄ T = P̄ 2π

ω = (0.3)

2π

30 = 0.063J

* 3. A block of mass m is connected to a spring, the other end of which is fixed. There isalso a viscous damping mechanism. The following observations have been made on thesystem:

(1) If the block is pushed horizontally with a force equal to mg, the static compressionof 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 differentialequation governing horizontal oscillations of the mass in terms of m,g, h and u.



(a) (1) tells us:

kxx=h

= mg

kh = mg

k = mg

h

(2) tells us:

bvv=u

= mg

bu = mg

b = mg

u

Using this in:

ma + bv + kx = 0

md2x

dt2 + b

dx

dt + kx = 0

gives:m

d2x

dt2 +

mg

u

dx

dt +

mg

h x = 0

d2x

dt2 +

g

u

dx

dt +

g

hx = 0

Comparing this with

d2x

dt2 + γ

dx

dt + ω2

0x = 0

gives us values for γ and ω0:

γ = g

u

ω20 = g

h ⇒ ω0 =

gh

Answer the following for the case that u = 3√

gh

(b) What is the angular frequency of the damped oscillations?

(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:

γ = g

u =

g

3√ gh =

1

3 g

hω20 =

g

h

(b) To find the angular frequency we use: ω2 = ω20 − γ 2

4 .

ω2 = ω20 −

γ 2

4

= g

h − 1

9

g

4h

= g

h − g

36h

= 35g

36h

ω =

35g

36h



(c) The energy decreases according to E (t) = E 0e−γt

E (t) = E 0e−γt

we need to find a τ st:

E (τ ) = E 0e−1 = E 0e−γτ

this tells us

1 = γτ

τ = 1

γ

= 3

h

g

So the time taken for the energy to decrease a factor of 1e

is t = 3

hg

s.

(d)

Q =

ω0

γ

=

gh

13

gh

= 113

= 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:x = C − v0

γ e−γt where γ = b

m

(b) At t = 0 the mass is at rest at x = 0. At this instant a driving force F = F 0 cos ωt isswitched 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 valuesof C and v0 from the conditions that x = 0 and dx

dt = 0 at t = 0

(a)

F = −bv

ma =−

bv

a = − b

mv = −γv

Now, we know that a = dvdt

:

dv

dt = −γv

1

vdv = −γdt

Integrating both sides gives 1

vdv =

−γdt = −γ

dt

ln(v) = −γt + D where D is a constantv = e−γt+D

v(t) = eDe−γt



Let v0 be the inital velocity at the time 0:

v(0) = eDe0 = v0

⇒ eD = v0

This gives us a final expression for v

v(t) = v0e−γt

To get an expression for x we use the fact that v = dxdt

dx

dt = v

dx

dt = v0e−γt

dx = v0e−γtdt

Integrating both sides gives dx =

v0e−γtdt = v0

e−γtdt

x = −v0

γ e−γt

+ C where C is a constant

⇒ x = C − v0γ

e−γt

(b)

ma + bv = F 0 cos ωt

md2x

dt2 + b

dx

dt = F 0 cos ωt

d2x

dt2 + γ

dx

dt =

F 0m

cos ωt

Looking at the steady state solution: x = A cos(ωt− δ ), we want to obtain expressions for A andδ .Using the complex exponential method, let


Aei(ωt−δ)

= Re(z)

z = Aei(ωt−δ)

dz

dt = iAωei(ωt−δ)

d2z


cos(ωt) = Re

eiωt


d2z

dt2 + γ

dz

dt =

F 0m

eiωt

−Aω2ei(ωt−δ) + γiAωei(ωt−δ) = F 0m

eiωt

−Aω2eiωte−iδ + Aiγωeiωte−iδ = F 0m

eiωteiωt

−Aω2e−iδ + Aiγωe−iδ = F 0m

−Aω2 + Aiγω = F 0m

eiδ

−Aω2 + Aiγω = F 0m

cos(δ ) + iF 0m

sin(δ )




−Aω2 = F 0m

cos(δ )

⇒ A = − F 0mω2

cos (δ )

Aγω =

F 0m sin(δ )

⇒ A = F 0γωm

sin (δ )

A = A ⇒− F 0mω2

cos (δ ) = F 0γωm

sin (δ )

− 1

ω2 cos (δ ) =

1

γ sin (δ )

sin(δ )

cos(δ ) = −γω

ω2 = −γ

ω

cos(δ ) = − ω

γ 2 + ω2

sin(δ ) = γ 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:

A = − F 0mω2

cos(δ )

= F 0mω2

ω γ 2 + ω2

= F 0

mω

γ 2 + ω2

(c) The general solution is give by x = C − v0γ

e−γt + A cos(ωt − δ )



We have the initial conditions: x(0) = dxdt

(0) = 0

x(t) = C − v0γ

e−γt + A cos(ωt − δ )

x(0) = C − v0γ

e0 + A cos(0− δ ) = 0

C −v0

γ + A cos(δ ) = 0dx

dt (t) = v0e−γt − ωA sin(ωt − δ )

dx

dt (0) = v0e0 − ωA sin(0− δ ) = 0

v0 + ωA sin(δ ) = 0

v0 = −ωA sin(δ )

= −ωA

γ

γ 2 + ω2

= −A

γω γ 2 + ω2

= − F 0

mω

γ 2 + ω2

γωA γ 2 + ω2

= − F 0γ

m(γ 2 + ω2)

And to find C :

C − v0γ

+ A cos(δ ) = 0

C = v0

γ −A cos(δ )

= −F 0γ

m(γ 2+ω2)

γ +

F 0

mω

γ 2 + ω2

ω γ 2 + ω2

= − F 0m(γ 2 + ω2)

+ F 0

m(γ 2 + ω2)

= 0

* 5. The graph shows the power resonance curve of a certain mechanical system when drivenby a force F 0 sin(ωt), where F 0 =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 energyof 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

Q = γ ≈ 2

Q = ω0

2 =

40

2 = 20

(b) The energy decreases according to E (t) = E 0e−

γt

E (t) = E 0e−γt

we need to find a τ st:

E (τ ) = E 0e−5 = E 0e−γτ

this tells us

5 = γτ

τ = 5

γ =

5

2 = 2.5

The time taken to complete one cycle is 2π

ω0 , so the time taken to complete n cycles is n2π

ω0 :

τ = 2.5 = n2π

ω0

⇒ n2π

ω0= 2.5

n2π

40 = 2.5

n(0.16) = 2.5

n = 2.5

0.16 = 15.6 ≈ 16

6. The figure shows the mean power input P̄ as a function of driving frequency for a masson a spring with damping. (Driving force = F 0 sin(ωt), where F 0 is held constant and ω isvaried.) 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 equationE = E 0e−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

Q = γ ≈ width of the power-resonance curve at half-height

ω0

Q = 1.02ω0 − 0.98ω0

= 0.04ω0

Q = ω0

0.04ω0

= 1

0.04 = 25

(b) From (a) we can just write down the value of γ :

γ = 0.04ω0

(c) The energy decreases according to the equation E = E 0e−γt , so the fraction of energy lost is E E 0

:

E E 0

= E 0e−γt

E 0= e−γt

= e−0.04ω0t

The time taken for one cycle is the perios T = 2πω0

, so the fraction of energy lost per cycle is

E

E 0= e

−0.04ω02π

ω0

= e−0.08πs

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 F 0 sin(ωt) is applied. In termsof the corresponding quantities for the original system, find the values of the following:

(d) The new resonant frequency ω

0.

(e) The new quality factor Q.

(f) The maximum mean power input P̄ m.

(g) The total energy of the system at resonance, E 0.

(d)

For the original system:

ω0 =

k

mFor the new system:

ω

0 =

k

m =

2k

m =√

2

k

m =√

2ω0

(e)


Q = ω0

γ For the new system:

Q = ω

0

γ =

√ 2

ω0

γ =

√ 2Q



(f)


P̄ max = QF 202mω0

For the new system:

P̄ max = Q

F 202mω

0

= √ 2QF 202m√

2ω0

= QF 202mω0

= P̄ max

(g) Originally we had E = E 0e−γt , therefore E 0 = Eeγt. As there is no k dependence here we seethat E 0 = Eeγt = E 0.

mit chapter 5 solutions

Documents