Quiz Samples for Chapter 15 General Physics I May 18, 2020...

General Physics IQuiz Samples for Chapter 15

OscillationsMay 18, 2020

Name: Department: Student ID #:

Notice

� +2 (−1) points per correct (incorrect) answer.

� No penalty for an unanswered question.

� Fill the blank ( ) with � (8) if the statement iscorrect (incorrect).

� Textbook: Walker, Halliday, Resnick, Principlesof Physics, Tenth Edition, John Wiley & Sons(2014).

15-1 Simple Harmonic Motion

1. (�) The frequency f of periodic, or oscillatorymotion is the number of oscillations per second. Inthe SI system, it is measured in hertz:

1 hertz = 1 Hz = 1 oscillation per second = 1 s−1.

2. (�) The period T is the time required for onecomplete oscillation, or cycle. It is related to thefrequency by

T =1

f.

3. In simple harmonic motion (SHM), thedisplacement x(t) of a particle from its equilibriumposition is described by the equation

x = xm cos(ωt+ φ).

(a) (�) xm is the amplitude of the displacement.

(b) (�) ωt+ φ is the phase of the motion.

(c) (�) φ is the phase constant.

(d) (�) The angular frequency ω is related to theperiod and frequency of the motion by

ω =2π

T= 2πf.

(e) (�) SHM velocity is

v = −ωxm sin(ωt+ φ).

(f) (�) SHM acceleration is

a = −ω2xm cos(ωt+ φ).

4. (�) The block–spring system is called a linearsimple harmonic oscillator (linear oscillator), wherelinear indicates that F is proportional to x to thefirst power (and not to some other power).

5. A particle with mass m that moves under theinfluence of a Hooke’s law restoring force given by

F = −kx

is a linear simple harmonic oscillator.

(a) (�) The displacement x satisfies thedifferential equation

x+ ω2x = 0,

where x =d2x

dt2, and the characteristic angular

frequency ω is defined by

ω =

√k

m.

(b) (�) The period of this oscillation is

T =2π

ω= 2π

√m

k.

(c) (�) The general solution for x(t) is

x(t) = c1 cosωt+ c2 sinωt,

where c1 and c2 are constants.

6. Consider the general solution for x(t) of the SHM:

x(t) = c1 cosωt+ c2 sinωt,


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(a) (�) The solution can be written in the form

x(t) =√c21 + c22

(c1√c21 + c22

cosωt

+c2√c21 + c22

sinωt

).

(b) (�) The solution can be written in the form

x(t) = xm(cosωt cosφ− sinωt sinφ)

= xm cos(ωt+ φ),

where the amplitude is

xm =√c21 + c22,

and the phase angle satisfies

tanφ = −c2c1.

7. A particle is in simple harmonic motion withperiod T . At time t = 0 it is halfway between theequilibrium point and the end point xm (> 0) of itsmotion, travelling toward the end point.

(a) (�) Consider the solution

x(t) = xm sin(ωt+ φ),

x(t) = ωxm cos(ωt+ φ),

x(t) = −ω2xm sin(ωt+ φ),

where the constant φ should be determined bymaking use of the initial conditions.Conditions state that

x(0) =xm2

= xm sinφ,

x(0) > 0.

(b) (�) The given constraints lead to thefollowing equation and inequality

sinφ =1

2, cosφ > 0.

The equation has the solutions φ = π/6 andφ = 5π/6. Taking into account the inequality,we find the solution φ = π/6.

(c) (�) The position, velocity, and accelerationare

x(t) = xm sin(ωt+

π

6

),

x(t) = ωxm cos(ωt+

π

6

),

x(t) = −ω2xm sin(ωt+

π

6

).

(d) (�) The next time it is at the same place is

ωt+π

6=

5π

6.

Thus

t =4π

6ω=T

3.

8. (�) Two identical undamped oscillators have thesame amplitude of oscillation only if they arestarted so the combination ω2x20 + v20 is the same,where x0 and v0 are the initial displacement andvelocity. And ω is the characteristic frequency ofthe oscillator.

15-2 Energy in Simple Harmonic Motion

1. Consider a particle in simple harmonic motionwithout friction. The solution for the displacementis

x(t) = xm cos(ωt+ φ),

where the amplitude xm and the phase constant φdo not depend on time and the angular frequencyω is

ω =

√k

m.

(a) (�) At any time, the kinetic energy is given by

K =1

2mv2 =

1

2x2m(mω2) sin2(ωt+ φ)

=1

2kx2m sin2(ωt+ φ).

(b) (�) At any time, the potential energy is givenby

U =1

2kx2 =

1

2kx2m cos2(ωt+ φ).

(c) (�) The total mechanical energy is conserved:

E =1

2kx2m =

1

2mx2mω

2.




2. A particle is in simple harmonic motion along thex axis. The amplitude of the motion is xm. Whenit is at x1, its kinetic energy is K1 and its potentialenergy (measured with U = 0 at x = 0) is U1.

(a) (�) The position, velocity, and accelerationare

x(t) = xm cos(ωt+ φ),

x(t) = −ωxm sin(ωt+ φ),

x(t) = −ω2xm cos(ωt+ φ).

(b) (�) The speed is

|x| = ω√x2m − x2.

(c) (�) The total mechanical energy E must beconserved.

U1 =1

2mω2x21,

K1 =1

2mω2(x2m − x21),

E = K1 + U1 =1

2mω2x2m.

(d) (�) At x = x2 = −12xm,

cos(ωt+ φ) = −1

2,

sin(ωt+ φ) = ±√

3

2,

U2 =1

4E,

K2 =3

4E.

3. A mass-spring system is oscillating with amplitudeA. At time t, the kinetic energy and the potentialenergy are the same. The potential energy, kineticenergy, and total mechanical energy are given by

U =1

2mω2A2 cos2(ωt+ φ),

K =1

2mω2A2 sin2(ωt+ φ),

E = K + U =1

2mω2A2.

(a) (�) Because U = K,

cos2(ωt+ φ) = sin2(ωt+ φ).

Thus

cos(ωt+ φ) = ± sin(ωt+ φ).

(b) (�) According to Pythagoras theorem, wehave one more constraint:

cos2(ωt+ φ) + sin2(ωt+ φ) = 1.

(c) (�) The solutions are

[cos(ωt+φ), sin(ωt+φ)] = (± 1√2,± 1√

2), (± 1√

2,∓ 1√

2).

(d) (�) The position and velocity are

x = ± 1√2A,

v = ± 1√2Aω,

or

x = ± 1√2A,

v = ∓ 1√2Aω.

15-3 An Angular Simple Harmonic Oscillator

1. A torsion pendulum consists of an objectsuspended on a wire. An angular version of asimple harmonic oscillator is associated with thetwisting of a suspension wire rather than theextension and compression of a spring as wepreviously had. The device is called a torsionpendulum, with torsion referring to the twisting.




(a) (�) The restoring torque of a linear torsionpendulum is given by

τ = −κθ,

where κ is the torsion constant of the wireand θ is the angular displacement from its restposition θ = 0.

(b) (�) The equation of motion is

τ = −κθ = Iθ = Id2θ

dt2,

where I is the rotational inertia of the objectabout the axis of rotation. We can rewrite theequation as

θ +κ

Iθ = 0.

This equation of motion is similar to that ofthe simple harmonic oscillator:

x+k

mx = 0.

(c) (�) The angular frequency ω of the oscillationis

ω =

√κ

I.

(d) (�) The period of the oscillation is

T =2π

ω= 2π

√I

κ.

15-4 Pendulums, Circular Motion

1. A simple pendulum consists of a rod of negligiblemass that pivots about its upper end, with aparticle (the bob) attached at its lower end.

(a) (�) If the rod swings through only smallangles, its motion is approximately simpleharmonic motion.

(b) (�) We choose i along the gravitationalacceleration g and j along the horizontaldirection (to the right) on the plane ofoscillation. The position vector of the bob is

r = Lr,

r = i cos θ + j sin θ,

� = −i sin θ + j cos θ.

Here, r the unit radial vector and � is the unittangential vector toward the direction ofincreasing θ.

(c) (�) Forces applied to the bob are

T = −T r,

Fg = mgi,

where T and Fg are the tension and thegravitational force.

(d) (�) The net force on the bob is

Fnet = −T r +mgi.

(e) (�) The net torque on the bob is

τnet = r × Fnet

= Lr× (−T r +mgi)

= mgLr× i,

where we have used the identity

r× r = 0.

(f) (�) i can be expressed as

i = −� sin θ + r cos θ.

(g) (�) There are useful identites involving crossproducts:

r× � = k,

i× j = k.

Here, k is the unit normal vector of the xyplane. In this case, it is coming out of thepaper.




(h) (�) The net torque on the bob is

τnet = r × Fnet

= −(mgL sin θ)r× �

= −(mgL sin θ)k.

(i) (�) The equation of motion is

τnet,z = −mgL sin θ = Id2θ

dt2,

where I is the rotational inertial of the bobabout the z axis,

I = mL2.

As a result, the equation of motion reducesinto

d2θ

dt2+mgL

Isin θ = 0.

(j) (�) If θ is small, then we can make a Taylor’sseries expansion and keep the leadingcontribution of θ as

sin θ = θ − θ3

3!+θ5

5!− · · · ≈ θ.

Then, the equation of motion reduces into

d2θ

dt2+mgL

Iθ = 0.

(k) (�) For a small amplitude case, the angularfrequency is

ω =

√mgL

I=

√g

L.

The period is

T =2π

ω= 2π

√I

mgL= 2π

√L

g.

2. A simple pendulum is suspended from the ceilingof an elevator. The elevator is acceleratingupwards with acceleration a.

(a) (�) The angular frequency of the oscillation,in terms of its length L, g and a is

ω′ =

√g + a

L.

(b) (�) The period of this pendulum, in terms ofits length L, g and a is

T ′ = 2π

√L

g + a.

(c) (�) If the elevator is accelerating downwardswith acceleration a, then the angularfrequency and the period are given by

ω′ =

√g − aL

,

T ′ = 2π

√L

g − a.

3. While a simple pendulum consists of a bob and amassless string, an arbitrary physical pendulum isa massive rigid body. The gravitational force Fg

acts at its center of mass C, at a distance h fromthe pivot point O.

Figure shows an arbitrary physical pendulumdisplaced to one side by angle θ.

(a) (�) The radius vector for the bob in a simplependulum is to be replaced with the vector−−→OC. Thus the length of the string in a simple

pendulum is to be replaced with h = |−−→OC|.

(b) (�) The equation of motion for the physicalpendulum is

d2θ

dt2+mgh

Isin θ = 0,

where I is the rotational inertia for the rigidbody about the pivot. From the parallel-axis




theorem, we can express I in terms of therotational inertia Icom about an axis passingthe center of mass and parallel to the pivot,m, and h as

I = Icom +mh2.

(c) (�) For a small amplitude case, the angularfrequency is

ω =

√mgh

I.

The period is

T =2π

ω= 2π

√I

mgh.

4. Simple harmonic motion is the projection ofuniform circular motion on a diameter of the circlein which the circular motion occurs.

(a) (�) The trajectory of a uniform circularmotion is

r(t) = rr

= r[i cos(ωt+ φ) + j sin(ωt+ φ)],

where r is a constant.

(b) (�) The projections onto x and y axes are,respectively, given by

x(t) = r cos(ωt+ φ),

y(t) = r sin(ωt+ φ).

The amplitudes of the oscillations for the xand y axes equal to

xm = ym = r.

(c) (�) Both x and y components satisfy theequation of motion for the simple harmonicoscillator,

x+ ω2x = 0,

y + ω2y = 0.

(d) (�) The velocity and the x and y componentsare

v(t) = rω�

= rω[−i sin(ωt+ φ) + j cos(ωt+ φ)],

x(t) = −rω sin(ωt+ φ),

y(t) = rω cos(ωt+ φ).

(e) (�) The acceleration and the x and ycomponents are

a(t) = −rω2r

= −rω2 [i cos(ωt+ φ) + j sin(ωt+ φ)],

x(t) = −rω2 cos(ωt+ φ),

y(t) = −rω2 sin(ωt+ φ).

15-5 Damped Simple Harmonic Motion

1. Let us consider a simple harmonic oscillator underthe presence of a damping force Fd that isproportional to the velocity v

Fd = −bv,

where b is the damping constant.

Equilibrium

(a) (�) The net force applied to the mass is

Fnet = −kx− bv.

(b) (�) The equation of motion is

mx+ bx+ kx = 0.

Because the equation has a single componentalong i, we find that

mx+ bx+ kx = 0.




2. Consider a damped simple harmonic oscillator

mx+ bx+ kx = 0.

Ifk

m>

(b

2m

)2

, then the solution is

x(t) = xme− b

2mt cos(ω′t+ φ),

where

ω′ =

√k

m−(

b

2m

)2

.

We verify this by substituting the solution to thedifferential equation explicitly.

(a) (�) The velocity is

v(t) = x(t)

= −xme−b

2mt

[b

2mcos(ω′t+ φ) + ω′ sin(ω′t+ φ)

].

Thus,

bv(t)

= xme− b

2mt

[− b2

2mcos(ω′t+ φ)− bω′ sin(ω′t+ φ)

].

(b) (�) The acceleration is

a(t) = x(t)

= xme− b

2mt

{[(b

2m

)2

− ω′2]

cos(ω′t+ φ)

+b

mω′ sin(ω′t+ φ)

}= xme

− b2m

t

[(b2

2m2− k

m

)cos(ω′t+ φ)

+b

mω′ sin(ω′t+ φ)

].

Thus

mx(t)

= xme− b

2mt[( b2

2m− k)

cos(ω′t+ φ)

+bω′ sin(ω′t+ φ)].

(c) (�) Substituting mx(t), bx(t), and kx(t) intothe equation of motion, we find that

mx+ bx+ kx = 0.

We call this case,k

m>

(b

2m

)2

, an

underdamped system.

3. [optional: This problem does not appear inthe quiz. Just for advanced students.] If

k

m=

(b

2m

)2


x(t) = [c1 + c2t] e− b

2mt,



(a) (�) kx(t) is

kx(t) =

[c1b

2

4m+c2b

2

4mt

]e−

b2m

t.

(b) (�) The velocity is

v(t) = x(t) =

[c2 −

c1b

2m− c2b

2mt

]e−

b2m

t.

Thus,

bx(t) =

[−c1b

2

2m+ c2b−

c2b2

2mt

]e−

b2m

t.

(c) (�) The acceleration is

a(t) = x(t)

=

[c1b

2

4m2− c2b

m+c2b

2

4m2t

]e−

b2m

t.

Thus

mx(t)

=

[c1b

2

4m− c2b+

c2b2

4mt

]e−

b2m

t.

(d) (�) Substituting mx(t), bx(t), and kx(t) intothe equation of motion, we find that

mx+ bx+ kx = 0.

We call this case,k

m=

(b

2m

)2

, the critical

damping.




(e) (�) The initial condition x(0) = x0 andx(0) = v0 can be imposed to determine theconstants c1 and c2 as

x(t) =

[x0 +

(v0 +

bx02m

)t

]e−

b2m

t.

4. [optional: This problem does not appear inthe quiz. Just for advanced students.] If

k

m<

(b

2m

)2


x(t) = c1e− b

2mt cosh(κt+ c2),

where c1 and c2 are constants and

κ =

√(b

2m

)2

− k

m.


(a) (�) The velocity is

v(t) = x(t)

= c1e− b

2mt

[− b

2mcosh(κt+ c2) + κ sinh(κt+ c2)

].

Thus,

bv(t) = x(t) = c1e− b

2mt

×[− b2

2mcosh(κt+ c2) + bκ sinh(κt+ c2)

].

(b) (�) The acceleration is

a(t) = x(t) = c1e− b

2mt

×[(

b2

4m2+ κ2

)cosh(κt+ c2)−

bκ

2msinh(κt+ c2)

]= c1e

− b2m

t

×[(

b2

2m2− k

m

)cosh(κt+ c2)−

bκ

msinh(κt+ c2)

].

Thus,

mx(t) = c1e− b

2mt

×[(

b2

2m− k)

cosh(κt+ c2)− bκ sinh(κt+ c2)

].

(c) (�) Substituting mx(t), bx(t), and kx(t) intothe equation of motion, we find that

mx+ bx+ kx = 0.

We call this case,k

m<

(b

2m

)2

, an

overdamped system.

5. Consider the damped harmonic oscillator,

mx+ bx+ kx = 0.

We restrict ourselves to the underdamped systemwith (

b

2m

)2

� k

m.

(a) (�) The energy loss per unit time can becomputed by

P = Fdv

= −bv2.

(b) (�) Because the sinusoidal factor oscillateswith the angular frequency

ω′ =

√k

m− b2

4m2≈ ω =

√k

m,

for each oscillation, it approximately takes

T =2π

ω′≈ 2π

√m

k.

(c) (�) The energy loss per each oscillation is

∆E =

∫ (n+1)T

nT[−bv2]dt

= −bx2m∫ (n+1)T

nTdte−

bmt

×[b

2mcos(ω′t+ φ) + ω′ sin(ω′t+ φ)

]2≈ −bx2mω2e−

bmt

∫ (n+1)T

nTdt sin2(ω′t+ φ)

= −bx2mω2e−bmt 1

2T.

(d) (�) For convenience, we take ∆t = T to findthat

∆E

∆t≈ −1

2bx2mω

2e−bmt.




(e) (�) The energy loss until time t is

E(t)− E(0) = −1

2bx2mω

2

∫ t

0e−

bmtdt

=1

2x2mmω

2(e−

bmt − 1

)=

1

2kx2m

(e−

bmt − 1

),

where we have made use of the identity,

k = mω2.

(f) (�) Because the initial energy is

E(0) =1

2kx2m,

the energy of the system at time t isapproximately

E(t) =1

2kx2me

− bmt.

15-6 Forced Oscillators and Resonance

1. (�) If an external driving force with angularfrequency ωd acts on an oscillating system withnatural angular frequency ω, the system oscillateswith angular frequency ωd. The velocity amplitudevm of the system is greatest when

ωd = ω,

a condition called resonance. The amplitude xm ofthe system is (approximately) greatest under thesame condition.

2. (�) An oscillator is driven by a sinusoidal force.The frequency of the applied force is independentof the natural frequency of the oscillator.

3. (�) A sinusoidal force with a given amplitude isapplied to an oscillator. To maintain the largestamplitude oscillation the frequency of the appliedforce should be determined from the maximumspeed desired.


Quiz Samples for Chapter 15 General Physics I May 18, 2020...

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