midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
Mechanics - Basic Physical Concepts
3
Math: Circle: 2 π r, π r2 ; Sphere: 4 π r2 , (4/3)
√ πr
2 −4 a c
−b±
b
2
Quadratic Eq.: a x + b x + c = 0, x =
2a
Cartesian and polar coordinates:
y
x = r cos θ, y = r sin θ, r2 = x2 + y 2 , tan θ = x
Trigonometry: cos α cos β + sin α sin β = cos(α − β)
α−β
sin α + sin β = 2 sin α+β
2 cos 2
α−β
cos α + cos β = 2 cos α+β
2 cos 2
sin 2 θ = 2 sin θ cos θ, cos 2 θ = cos2 θ − sin2 θ
1 − cos θ = 2 sin2 2θ , 1 + cos θ = 2 cos2 2θ
~ = (Ax , Ay ) = Ax ı̂ + Ay ̂
Vector algebra: A
~ =A
~+B
~ = (Ax + Bx , Ay + By )
Resultant:
R
~·B
~ = A B cos θ = Ax Bx + Ay By + Az Bz
Dot: A
Cross product: ı̂ × ̂ = k̂, ̂ × k̂ = ı̂, k̂ × ı̂ = ̂
¯
¯
¯ ı̂
̂
k̂ ¯
¯
¯
~
~
~
¯
C = A × B = ¯ Ax Ay Az ¯¯
¯
¯ B
x By Bz
C = A B sin θ = A⊥ B = A B⊥ , use right hand rule
d xn = n xn−1 ,
d
1
Calculus: dx
dx ln x = x ,
d sin θ = cos θ,
d cos θ = − sin θ,
d
dθ
dθ
dx const = 0
Measurements
Dimensional analysis: e.g.,
2
F = m a → [M ][L][T ]−2 , or F = m vr → [M ][L][T ]−2
PN
PN
Summation:
i=1 (a xi + b) = a i=1 xi + b N
Motion
One dimensional motion: v = ddts , a = ddtv
s −s
v −v
Average values: v̄ = tff −tii , ā = tff −tii
One dimensional motion (constant acceleration):
v(t) : v = v0 + a t
s(t) : s = v̄ t = v0 t + 12 a t2 , v̄ = v02+v
v(s) : v 2 = v02 + 2 a s
Nonuniform acceleration: x = x0 + v0 t + 12 a t2 +
1 j t3 + 1 s t4 + 1 k t5 + 1 p t6 + . . ., (jerk, snap,. . .)
6
24
120
720
ttrip
v0y
2 = g
1
2
h = 2 g tf all , R = vox ttrip
2
Circular: ac = vr , v = 2 Tπ r , f = T1 (Hertz=s−1 )
q
Curvilinear motion: a = a2t + a2r
Relative velocity: ~v = ~v ′ + ~u
Projectile motion: trise = tf all =
Law of Motion and applications
Force: F~ = m ~a, Fg = m g, F~12 = −F~21
2
Circular motion: ac = vr , v = 2 Tπ r = 2 π r f
Friction: Fstatic ≤ µs N
Fkinetic = µk N
P
Equilibrium (concurrent forces):
~
i Fi = 0
Energy
Work (for all F): ∆W = WAB = WB − WA
1
RB
F~ · d~s (in Joules)
Fk s = F s cos θ = F~ · ~s → A
Effects due to work done: F~ext = m ~a − F~c − f~nc
Wext |A→B = KB − KA + UB − UA + Wdiss |A→B
RB
Kinetic energy: KB −KA = A
m ~a ·d~s, K = 12 m v 2
R
K (conservative F~ ): U − U = − B F~ · d~s
B
A
A
Ugravity = m g y,
Uspring = 12 k x2
∂
U
From U to F~ : Fx = − ∂x , Fy = − ∂∂yU , Fz = − ∂∂zU
Fgravity = − ∂∂yU = −m g,
U = −k x
Fspring = − ∂∂x
2
∂
U
∂
U
Equilibrium: ∂x = 0, ∂x2 > 0 stable, < 0 unstable
W = F v = F v cos θ = F
~ · ~v (Watts)
Power: P = ddt
k
Collision
Rt
Impulse: I~ = ∆~
p = p~f − p~i → tif F~ dt
Momentum: p~ = m ~v
x1 +m2 x2
Two-body: xcm = m1m
1 +m2
pcm ≡ M vcm = p1 + p2 = m1 v1 + m2 v2
Fcm ≡ F1 + F2 = m1 a1 + m2 a2 = M acm
K1 + K2 = K1∗ + K2∗ + Kcm
Two-body collision: p~i = p~f = (m1 + m2 ) ~vcm
vi′ = vi∗′ + vcm
vi∗ = vi − vcm ,
Elastic: v1 − v2 = −(v1′ − v2′ ),
vi∗′ = −vi∗ , vi′ = 2 vcm − vi
R
P
~r dm
m ~r
Many body center of mass: ~rcm = P i i = R
mi
mi
P
p
Force on cm: F~ext = d~
=
M~
a
,
p
~
=
p
~
cm
i
dt
Rotation of Rigid-Body
Kinematics: θ = rs , ω = vr , α = art
R
P
Moment of inertia: I =
mi ri2 = r2 dm
1
1
2
Idisk = 2 M R , Iring = 2 M (R12 + R22 )
1 M ℓ2 , I
1
2
2
Irod = 12
rectangle = 12 M (a + b )
Isphere = 25 M R2 , Ispherical shell = 23 M R2
I = M (Radius of gyration)2 , I = Icm + M D2
Kinetic energies: Krot = 12 I ω 2 , K = Krot + Kcm
Angular momentum: L = r m v = r m ω r = I ω
Torque: τ = ddtL = m ddtv r = F r = I ddtω = I α
Wext = ∆K +∆U +Wf ,
K = Krot + 12 m v 2 ,
P =τω
Rolling, angular
momentum
and
´
³
´
³ torque
Ic + M v 2
Rolling: K = 12 Ic + M R2 ω 2 = 12 R
2
~
Angular momentum: L = ~r × p~, L = r⊥ p = I ω
~
Torque: ~τ = d L = ~r × d~p = ~r × F~ , τ = r F = I α
dt
dt
1 dL = τ = mgh
Gyroscope: ωp = ddtφ = L
L
Iω
dt
Static equilibrium
P
P~
~τi = 0
Fi = 0, about any point
+mB ~rBcm
Subdivisions: ~rcm = mA ~rAcm
mA +mB
Elastic modulus = stress/strain
stress: F/A
strain: ∆L/L, θ ≈ ∆x/h, −∆V /V
⊥
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
Gravity
F~21 = −G m12m2 r̂12 ,
r12
for r ≥ R,
g(r) = G M
r2
G = 6.67259 × 10−11 N m2 /kg2
Rearth = 6370 km, Mearth = 5.98 × 1024 kg
³ ´2
2
Circular orbit: ac = vr = ω 2 r = 2Tπ r = g(r)
U = −G mrM ,
M
E = U + K = −Gm
2r
r0
r2 = 1−ǫ
Fluid mechanics
Pascal: P = FA⊥1 = FA⊥2 , 1 atm = 1.013 × 105 N/m2
1
2
Archimedes: B = M g, Pascal=N/m2
P = Patm + ρ g h, with P = FA⊥ and ρ = m
V
R
R
F = P dA −→ ρ g ℓ 0h (h − y) dy
Continuity equation: A v = constant
Bernoulli: P + 21 ρ v 2 + ρ g y = const,
P ≥0
Oscillation motion
f = T1 , ω = 2Tπ
2
2
S H M: a = ddt2x = −ω 2 x, α = ddt2θ = −ω 2 θ
x = xmax cos(ω t + δ), xmax = A
v = −vmax sin(ω t + δ), vmax = ω A
a = −amax cos(ω t + δ) = −ω 2 x, amax = ω 2 A
E = K + U = Kmax = 12 m (ω A)2 = Umax = 21 k A2
Spring: m a = −k x
Simple pendulum: m aθ = m α ℓ = −m g sin θ
Physical pendulum: τ = I α = −m g d sin θ
Torsion pendulum: τ = I α = −κ θ
Wave motion
Traveling waves: y = f (x − v t), y = f (x + v t)
In the positive x direction: y = A sin(k x − ω t − φ)
λ
T = f1 , ω = 2Tπ , k = 2λπ , v = ω
k = T
q
Along a string: v = F
µ
fixed end: phase inversion
open end: same phase
General: ∆E = ∆K + ∆U = ∆Kmax
1 ∆m
2
P = ∆E
∆t = 2 ∆t (ωA)
∆m ∆x
∆m
Waves: ∆m
∆t = ∆x · ∆t = ∆x · v
P = 21 µ v (ω A)2 , with µ = ∆m
∆x
∆m ∆A ∆r
∆m
Circular: ∆m
∆t = ∆A · ∆r · dt = ∆A · 2 π r v
∆m
2
Spherical: ∆m
∆t = ∆V · 4 π r v
v=
q
Sound
B,
ρ
s = smax cos(k x − ω t − φ)
1
2
Intensity: I = P
A = 2 ρ v (ω smax )
I
Intensity level: β = 10 log10 I , I0 = 10−12 W/m2
0
Plane waves: ψ(x, t) = c sin(k x − ω t)
Circular waves: ψ(r, t) = √c sin(k r − ω t)
Spherical: ψ(r, t) = rc sin(k r − ω t)
L
⊥
⊥
ii) L = r m ∆r
−→ ∆A = 21 r ∆r
const.
∆t = 2 m =
³ 2´
³∆t ´2 ∆t
r
1
4π
2
π
a
M
3
2
1 +r2
iii) G a2 =
a, a =
2 , T = GM r
T
Escape kinetic energy: E = K + U (R) = 0
Reflection of wave:
∂s
∆P = −B ∆V
V = −B ∂x
∆Pmax = B κ smax = ρ v ω smax
∆m A ∆x
Piston: ∆m
∆t = ∆V · ∆t = ρ A v
r
2
F = − ddrU = −m G M
= −m vr
r2
Kepler’s Laws of planetary motion:
r0
0
i) elliptical orbit, r = 1−ǫrcos
θ r1 = 1+ǫ ,
2
′
v
Doppler effect: λ = v T , f0 = T1 , f ′ = λ
′
′
Here v = vsound ± vobserver , is wave speed relative
to moving observer and λ′ = (vsound ± vsource )/f0 ,
detected wave length established by moving source of
frequency f0 . freceived = fref lected
Shock waves: Mach Number= vvsource
= sin1 θ
sound
Superposition of waves
Phase difference: sin(k x − ωt) + sin(k x − ω t − φ)
Standing waves: sin(k x − ω t) + sin(k x + ω t)
Beats: sin(kx − ω1 t) + sin(k x − ω2 t)
Fundamental modes: Sketch wave patterns
λ
String: λ
2 = ℓ, Rod clamped middle: 2 = ℓ,
Open-open pipe: λ
2 = ℓ,
Open-closed pipe: λ
4 =ℓ
Temperature and heat
Conversions: F = 95 C + 32◦ ,
K = C + 273.15◦
Constant volume gas thermometer: T = a P + b
Thermal expansion: α = 1ℓ ddTℓ , β = V1 ddTV
∆ℓ = α ℓ ∆T , ∆A = 2 α A ∆T , ∆V = 3 α V ∆T
Ideal gas law:
P V = nRT = N kT
R = 8.314510 J/mol/K = 0.0821 L atm/mol/K
k = 1.38 × 10−23 J/K, NA = 6.02 × 1023 , 1 cal=4.19 J
Calorimetry: ∆Q = c m ∆T, ∆Q = L ∆m
R
First law: ∆U = ∆Q − ∆W , W = P dV
−H ℓi
∆T
Conduction: H = ∆Q
∆t = −k A ∆ℓ , ∆Ti = A ki
Stefan’s law: P = σ A e T 4 , σ = 5.67 × 10−8 m2WK 4
Kinetic theory of gas
2
m vx
x
F = ∆p
∆t = d
N 2
Pressure: P = NAF = mVN vx2 = m
3V v
K
1
P = 23 N
V K, K x = 3 = 2 k T , T = 273 + Tc ,
P V = N k T , n = N/NA , k = 1.38×10−23 J/K,
NA = 6.02214199 × 1023 #/kg/mole
Constant V:
∆Q = ∆U = n CV ∆T
Constant P:
∆Q = n CP ∆T
Ideal gas: ∆px = 2 m vx ,
C
γ = CP , CP − CV = R
V
CV = d2 R, for transl.+rot+vib, d = 3 + 2 + 2
Adiabatic expansion: P V γ = constant
t
1
Mean free path: ℓ = (v vrms
= √ 12
)
t π d2 n
rel rms
V
2 π d nV
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
Question 1, chap 16, sect 4.
part 1 of 1
10 points
3
6.
The diagrams below show different standing waves on a 90 cm string.
Which of these waves has a 60 cm wavelength?
7.
1.
8.
2.
9.
3.
Explanation:
A standing wave on a string has nodes
where the string does not vibrate; these nodes
are spaced half-wavelength from each other:
4.
x0 = 0,
x1 =
λ
,
2
x2 = 2 ×
λ
,
2
λ
λ
, x4 = 4 × , . . . .
2
2
A fixed end of the string cannot vibrate, so it
must be a node of the standing wave. For the
string fixed at both ends, both ends must be
nodes of the wave, hence
x3 = 3 ×
5.
correct
λ
, n = 1, 2, 3, . . . ,
(1)
2
where n is the number of vibrating segments
of the string. More precisely, the wave on
the string has n antinodes and n + 1 nodes:
One node at each end of the string, plus n − 1
nodes in the middle.
The wave in question has wavelength λ =
60 cm on a string of length L = 90 cm, hence
L = n×
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
according to eq. (1),
n =
4
v
5.
L
90 cm
=
= 3,
λ/2
(60 cm)/2
which means the wave has 3 antinodes and 4
nodes: One at each end, and 2 in the middle.
In other words, it looks like
Explanation:
Consider the wave pattern image reflected
about the rigid hook on the wall.
v
.
Question 2, chap 16, sect 99.
part 1 of 1
10 points
The figure below shows a complex wave
pattern on a string moving towards a rigid
hook at the wall on the right. After some
time, the wave is reflected from the wall.
v
After the time it takes for the wave to be
reflected from the wall, this image is the wave
pattern traveling to the left along the string.
Note: Reflection about a point (hook) is the
same as reflection about the y-axis (wall) followed by reflection about the x-axis (string).
The leading part of the wave must remain in
front and the wave is flipped over.
v
v
v
This is the third wave pattern of four possible wave patterns presented by this question.
Select the wave pattern for the reflected wave.
v
correct
1.
Question 3, chap 17, sect 2.
part 1 of 1
10 points
A skyrocket explodes 100 m above the
ground. Three observers are spaced 93 m
apart, with observer A directly under the
point of the explosion.
v
2.
100 m
v
3.
A
B
93 m
4.
C
93 m
Find the ratio of the sound intensity heard
by observer A to that heard by observer B.
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
5
The figure is not drawn to scale. For practical
purposes, you may treat each observer as a
point on the ground.
Correct answer: 1.8649.
Explanation:
2.6 kg
Let : h = 100 m
d = 93 m .
and
What is the wave speed when the suspended
mass is 3 kg?
Correct answer: 29.0027.
rA = 100 m
Source
b
Explanation:
rC
rB
B
b
b
b
A
C
93 m
93 m
The intensity at a distance r from the
source is
P
I=
,
4 π r2
and distances from the source to points A and
B are
rA = h and
p
rB = h2 + d2 ,
so
IA
r2
h2 + d2
= B2 =
IB
h2
rA
(100 m)2 + (93 m)2
=
(100 m)2
Let : m1 = 2.6 kg ,
m2 = 3 kg , and
v1 = 27 m/s .
The linear density is
s
F
v1 =
, so
µ
F
µ= 2
v1
m1 g
= 2
v1
(2.6 kg) (9.8 m/s2 )
=
(27 m/s)2
= 0.034952 kg/m .
The velocity is
s
v2 =
v1 =
= 1.8649 .
=
Question 4, chap 16, sect 2.
part 1 of 1
10 points
Tension is maintained in a string as in the
figure. The observed wave speed is 27 m/s
when the suspended mass is 2.6 kg .
The acceleration of gravity is 9.8 m/s2 .
r
s
F
,
µ
(1)
so
m2 g
µ
(2)
(3 kg) (9.8 m/s2 )
0.034952 kg/m
= 29.0027 m/s .
Alternate Solution: Plugging µ from Eq.
1 into Eq. 2, we have
r
m2 g
(2)
v1 =
µ
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
s
or
m2 g v12
1
=
k A2 = m v 2 + k A2
m1 g
4
r
m2
so
(3)
= v1
m1
s
3 k A2
m
=
(3 kg)
4 v2
= (27 m/s)
(2.6 kg)
(3) (3.4 N/m) (0.0483 m)2
=
(4) (0.235 m/s)2
= 29.0027 m/s .
= 0.107721 kg .
Question 5, chap 15, sect 3.
part 1 of 1
10 points
A block of unknown mass is attached to
a spring of spring constant 3.4 N/m and
undergoes simple harmonic motion with an
amplitude of 4.83 cm. When the mass is
halfway between its equilibrium position and
the endpoint, its speed is measured to be
v = 23.5 cm/s.
Calculate the mass of the block.
Correct answer: 0.107721.
Explanation:
Basic Concepts Energy conservation: If
K is kinetic energy and U is potential energy,
6
Question 6, chap 17, sect 4.
part 1 of 1
10 points
A police car is traveling at a speed vc to the
right. A truck is traveling at a speed vt to the
right. The frequency of the siren on the police
car is fc .
The speed of sound in air is va . Let vt be
the speed of the observer in the truck, and
vc the speed of the source, the police car.
vc
vt
Police
Truck
Ki + Ui = Kf + Uf
Kinetic energy of particle with mass m and
speed v:
1
K = m v2
2
Mass m on spring with constant k:
r
k
ω=
m
and potential energy of a spring at displacement x:
1
U = k x2
2
Period
2π
T =
ω
Solution: Call the maximum displacement
(amplitude) A. The halfway displacement is
A/2. Energy conservation requires
2
1
1
A
1
2
2
0 + kA = mv + k
2
2
2
2
What is the frequency ft heard by an observer in the moving truck?
va − vt
va + vc
va + vt
2. ft =
va − vc
va − vt
3. ft =
va − vc
va + vt
4. ft =
va + vc
1. ft =
fc
fc
fc correct
fc
Explanation:
The Doppler shifted frequency f ′ heard in
the truck is
f′ =
va ± vo
f,
va ∓ vs
(1)
where va is the speed of sound in air, vo is
the speed of the observer, and vs is the speed
of the source. The upper sign is used when
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
the relative velocities are toward one another,
and vice versa.
The relative velocity of the observer is away
from the source so the lower sign is used in
the numerator (± → −), and the relative
velocity of the source is toward the observer
so the upper sign is used in the denominator
(∓ → −). Therefore Eq. 2 becomes
ft =
va − vt
fc .
va − vc
Question 7, chap 16, sect 1.
part 1 of 1
10 points
Sound in air can best be described as which
of the following type of wave?
1. Electromagnetic
2. Torsional
v=
a=
7
dx
= ω A cos(ω t + φ)
dt
dv
= −ω 2 A sin(ω t + φ)
dt
The basic concepts above are enough to solve
the problem. Just use the formula for a
obtained by differentiating x twice, where
t = 1 s:
a = −ω 2 A sin(ω t + φ)
= −(3.9 rad/s)2 (5.7 m)
× sin[(3.9 rad/s) (1 s) + (1.0472 rad)]
= 84.3179 m/s2 .
The phase φ (given in radians) incorporates
the initial condition where the body started
(t = 0), meaning it started at
3. Polarized
4. Longitudinal correct
5. Transverse
Explanation:
A sound wave in the air is propagated by
the oscillation of air molecules. It is best
described as a longitudinal wave.
Question 8, chap 15, sect 2.
part 1 of 1
10 points
A body oscillates with simple harmonic motion along the x-axis. Its displacement varies
with time according to the equation,
x(t) = A sin(ω t + φ).
If A = 5.7 m, ω = 3.9 rad/s, and φ =
1.0472 rad, what is the magnitude of the acceleration of the body at t = 1 s?
Note: The argument of the sine function is
given here in radians rather than degrees.
Correct answer: 84.3179.
Explanation:
x = A sin(ω t + φ)
x0 = A sin φ
= (5.7 m) sin(1.0472 rad)
= 4.93634 m
and it is now at
x = A sin(ω t + φ)
= (5.7 m)
× sin[(3.9 rad/s) (1 s) + (1.0472 rad)]
= −5.54359 m .
(These two last facts are not needed to solve
the problem but clarify the physical picture.)
Question 9, chap 16, sect 2.
part 1 of 1
10 points
A harmonic wave
y = A sin[k x − ω t − φ] ,
where A = 1 meter, k has units of m−1 , ω
has units of s−1 , and φ has units of radians, is
plotted in the diagram below.
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
t = 0)
At the time t = 0
y = A sin k x − φ
2π
= A sin
x−φ
λ
2π
x−φ
= A sin
9m
A (meters)
+1
12
18
Which wave function corresponds best to
the diagram?
4π
2π
x−ωt−
1. y = A sin
3m
3
2π
5π
2. y = A sin
x−ωt−
9m
3
2π
4π
3. y = A sin
x−ωt−
15 m
3
2π
2π
4. y = A sin
x−ωt−
15 m
3
2π
1π
x−ω t−
cor5. y = A sin
9m
3
rect
2π
5π
6. y = A sin
x−ωt−
3m
3
2π
1π
x−ωt−
7. y = A sin
3m
3
4π
2π
x−ωt−
8. y = A sin
9m
3
2π
1π
9. y = A sin
x−ωt−
15 m
3
2π
2π
10. y = A sin
x−ωt−
3m
3
Explanation:
From the diagram of the wave function the
wave-length λ = 9 m (6 horizontal scale divisions of 1.5 m each, see diagram below).
Notice: Since one wave-length is 2 π raπ
2π
=
dians, each horizontal division is
6
3
radians.
The given wave function (sine function with
(gray curve in diagram below), therefore
φ=1
π
1π
=
radians .
3
3
Checking the wave function y at x = 0, we
have agreement with the diagram below
1π
y = sin 0 −
= −0.866025 m .
3
At the time t = 0
+1
A (meters)
6
(dark curve in diagram below) is shifted 1
divisions to the right (negative phase shift) of
a no-phase-shift sine function
2π
x
y = sin
9m
−1
φ
−1
x (meters)
8
9
x (meters)
18
λ=9
Therefore, the wave function is
1π
2π
x−ωt−
y = A sin
.
9m
3
keywords:
Question 10, chap 18, sect 5.
part 1 of 1
10 points
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
9
ρwater
mbl g
A light spring with a spring constant of
=
−1
16.6 N/m rests vertically on the bottom of a
ρbl
k
large beaker of water, as shown in (a).
3
1000 kg/m
2
=
−1
The acceleration of gravity is 9.81 m/s .
644.1 kg/m3
A 0.00473 kg block of wood with a density
(0.00473 kg) (9.81 m/s2 )
of 644.1 kg/m3 is connected to the spring, and
×
16.6 N/m
the mass-spring system is allowed to come
to static equilibrium, as shown in (b). The
= 0.00154453 m .
magnitude of the force pulling the spring back
to its unstretched position equals k ∆x.
m
Question 11, chap 17, sect 3.
part 1 of 1
10 points
∆x
This picture shows the displacements S of
the air molecules in a traveling sound wave as
a function of distance, x.
Sound Wave
k
k
(a)
(b)
+A
How much does the spring stretch?
Correct answer: 0.00154453.
S
Explanation:
Basic Concepts:
Fnet = FB − Fg,bl − Fspring = 0
FB = mf luid g
Fg = m g
Fspring = k ∆x
m
ρ=
V
Given:
k = 16.6 N/m
mbl = 0.00473 kg
ρbl = 644.1 kg/m3
ρwater = 1000 kg/m3
g = 9.81 m/s2
−A
λ/2
3 λ/2
0
λ
2λ
Which of the following tubes, closed at the
left end and closed at the right end, is closest
to the right length ℓ so as to resonate at its
fundamental frequency when placed in this
sound wave?
1. ℓ =
2. ℓ = 2 λ
1
3. ℓ = λ
2
correct
Solution:
ρwater V g − mbl g − k ∆x = 0
ρwater
∆x =
mbl
g − mbl g
ρbl
k
1
λ
4
4. ℓ =
3
λ
4
5. ℓ = λ
6. ℓ =
3
λ
8
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
3
λ
2
1
8. ℓ = λ
8
10
7. ℓ =
hℓ
0.83 cm
light liquid
2.5 g/cm3
0.93 cm
heavy liquid
8.9 g/cm3
Explanation:
The tube is closed at the left end requiring
an node in the wave form. The tube is closed
at the right end requiring an node in the wave
form.
The displaced between the node on the left
and the node on the right is shown in the
figure below. This correspondance can be
1
represented by the distance between ℓ = λ
2
and ℓ = λ as shown.
Sound Wave
+A
If the equilibrium configuration of the tube
is as shown in the figure, with a difference
in the height of the heavy liquid of 0.93 cm,
determine the value of the difference in height
of the light liquid hℓ .
Correct answer: 2.3808.
Explanation:
Let :
ρℓ = 2.5 g/cm3 ,
ρh = 8.9 g/cm3 ,
hh = 0.93 cm , and
h = 0.83 cm , not required.
Also, let h be the height of the light liquid
column added to the right side of the U-tube.
Consider the pressure at the elevation of the
light-heavy liquid interface in the left column
and at the same elevation in the right column.
By Pascal’s Principle, the absolute pressure
at that elevation is the same in both columns.
We have,
S
−A
λ/2
3 λ/2
0
λ
2λ
So the length of a tube to produce the
1
fundamental resonance should be ℓ = λ .
2
Third of four versions.
Question 12, chap 18, sect 4.
part 1 of 1
10 points
A U-tube of constant cross-sectional area,
open to the atmosphere, is partially filled with
a heavy liquid with density 8.9 g/cm3 . A light
liquid with density 2.5 g/cm3 is then poured
into both arms.
Pright = Patm + ρh g hh + ρℓ g h
and
Plef t = Patm + ρℓ g (hh + h + hℓ ) .
From Pascal’s Principle, Plef t = Pright
ρℓ (hh + hℓ ) = ρh (hh ) , or
ρh
hℓ =
− 1 hh
ρℓ
8.9 g/cm3
− 1 hh
=
2.5 g/cm3
= 2.3808 cm .
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
Question 13, chap 17, sect 2.
part 1 of 1
10 points
Middle C has a speed of 1500 m/s in water
and 340 m/s in air.
Does it have a longer or shorter wavelength
in water than in air and why?
11
P1
P2 1.4 atm
2.13 m/s
0.15 m
v2
1. Shorter wavelength in water because the
speed of sound is greater in water than in air
7.49 m
0.05 m
2. Longer wavelength in water because water
is denser than air
Applying Bernoulli’s principle, what is the
pressure P1 at the entrance end of the pipe?
Correct answer: 286099.
3. Longer wavelength in water because the
speed of sound is greater in water than in air
correct
Explanation:
Applying Bernoulli’s principle to the fluid
flow at the entrance and exit of the pipe gives
4. Shorter wavelength in water because water is denser than air
P1 + ρ g y 1 +
1 2
1
ρ v1 = P2 + ρ g y2 + ρ v22
2
2
Explanation:
P1 = P2 + ρ g (y2 − y1 ) +
wavelength =
velocity
frequency
The velocity of sound in water is larger than
in air. The frequency doesn’t change, so the
wavelength in water is larger than in air.
Question 14, chap 18, sect 6.
part 1 of 1
10 points
Note: Patm = 101300 Pa/atm. The viscosity of the fluid is negligible and the fluid is
incompressible..
A liquid of density 1335 kg/m3 flows with
speed 2.13 m/s into a pipe of diameter 0.15 m.
The diameter of the pipe decreases to 0.05 m
at its exit end. The exit end of the pipe is
7.49 m lower than the entrance of the pipe,
and the pressure at the exit of the pipe is
1.4 atm.
The acceleration of gravity is 9.8 m/s2 .
1
ρ (v2 2 − v1 2 ) .
2
We also have y2 − y1 = −h, since the entrance
height y1 is greater than the exit height y2 .
Therefore
P1 = P2 − ρ g h +
1
ρ [v22 − v1 2 ]
2
= (141820 Pa)
− (1335 kg/m3 ) (9.8 m/s2 ) (7.49 m)
1
+ (1335 kg/m3 )
2
× [(19.17 m/s)2 − (2.13 m/s)2 ]
= 286099 Pa .
Question 15, chap 18, sect 4.
part 1 of 1
10 points
The small piston of a hydraulic lift has a
cross-sectional area of 9 cm2 and the large
piston has an area of 14 cm2 , as in the figure
below.
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
77 kN
F
area
9 cm2
14 cm2
12
2. By making a string (a) shorter, (b)
tighter, and (c) lighter.
3. By making a string (a) longer, (b) tighter,
and (c) more massive.
4. By making a string (a) longer, (b) looser,
and (c) more massive. correct
5. By making a string (a) longer, (b) looser,
and (c) lighter.
What force F must be applied to the small
piston to raise a load of 77 kN?
Correct answer: 49500.
Explanation:
Let : A1
A2
W
F
= 9 cm2 ,
= 14 cm2 ,
= 77 kN , and
=F.
According to Pascal’s law, the pressure exerted on A1 must be equal to the one exerted
F
on A2 . The pressure P1 =
must be equal
A1
W
due to the load.
to the pressure P2 =
A2
F
W
=
, so
A1
A2
A1
W
F =
A2
(9 cm2 )
=
(77000 N)
(14 cm2 )
= 49500 N .
Question 16, chap 16, sect 2.
part 1 of 1
10 points
Explain how you can lower the pitch of a
tone on a guitar by altering (a) the length of
the string, (b) the tension of the string, or (c)
the thickness or the mass of the string.
1. By making a string (a) shorter, (b) looser,
and (c) more massive.
Explanation:
A low pitch will be produced when a guitar
string is (a) lengthened, (b) loosened so that
tension is reduced, and (c) made more massive, usually by windings of wire around the
string. That’s why bass strings are thick.
Question 17, chap 16, sect 1.
part 1 of 1
10 points
Earthquakes produce two kinds of seismic waves: The longitudinal primary waves
(called P waves) and the transverse secondary
waves (called S waves).
Both S waves and P waves travel through
the Earth’s crust and mantle, but do so at different speeds; the P waves are always faster
than the S waves, but their exact speeds depend on depth and location.
For the purpose of this exercise, we assume
the P wave’s speed to be vP = 8370 m/s
while the S waves travel at a slower speed of
vS = 4030 m/s.
Suppose a seismic station detects a P wave
and then ∆t = 89.3 s later detects an S wave.
How far away is the earthquake center?
Correct answer: 694.052.
Explanation:
Suppose the earthquake happens at time
t = 0 at some distance d. The P wave and
the S wave are both emitted at the same
time t = 0, but they arrive at different times,
respectively tP = d/vP and tS = d/vS . The
S wave is slower, so it arrives later than the
P wave, the time difference being
d
d
d(vP − vS )
∆t =
−
=
.
vS
vP
vP vS
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
Consequently, given this time difference and
the two waves’ speeds vP and vS , we find the
earthquake center being
Finally, the period of the oscillation follows
from the angular frequently,
vP vS ∆t
vP − vS
(8370 m/s) (4030 m/s) (89.3 s)
=
(8370 m/s) − (4030 m/s)
d=
= 694.052 km .
away from the seismic station.
Question 18, chap 15, sect 1.
part 1 of 1
10 points
Hint: Write down equations for x(t) and
v(t) and use sin2 + cos2 = 1 to calculate ω.
A mass attached to a spring executes simple harmonic motion in a horizontal plane
with an amplitude of 2.97 m. At a point
1.5741 m away from the equilibrium, the mass
has speed 0.554 m/s.
What is the period of oscillation of the
mass?
Correct answer: 28.5641.
Explanation:
There are several ways to solve this problem — for example, one may use the energy
conservation — but the simplest solution uses
nothing but the equation of simple harmonic
motion
x(t) = A sin(ω t) ,
13
T =
2π
= 28.5641 s .
ω
Question 19, chap 18, sect 5.
part 1 of 1
10 points
Assume: The buoyant force of the air is
negligible.
Given: The density of the oil is 874 kg/m3 .
The density of the water salt mixture is
1153 kg/m3 .
A rectangular parallelepiped block of uniform density floats in a container which contains oil and salt water as shown. The block
sticks up above the oil by a distance 20 cm.
The oil thickness is 30 cm. The block’s depth
in the salt water is 20 cm. The horizontal area
of the block is 0.03 m2 .
20 cm
30 cm
M
and its time derivative
20 cm
dx
v(t) =
= A ω cos(ω t) .
dt
For any angle [such as the phase angle ω t]
sin2 (ω t) + cos2 (ω t) = 1 ,
hence
x 2
+
A
and therefore
v 2
ω
v 2
= 1,
Aω
= A 2 − x2 .
Consequently,
ω=√
v
= 0.219968 s−1 .
2
A − x2
Calculate the mass of the block.
Correct answer: 14.784.
Explanation:
Let :
d1 = 20 cm = 0.2 m ,
d2 = 30 cm = 0.3 m ,
d3 = 20 cm = 0.2 m ,
A = 0.03 m2 ,
and
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
The mass of the salt water displaced is
M 3 = ρ w d3 A
= (1153 kg/m3 ) (0.2 m) (0.03 m2 )
= 6.918 kg .
The wave speed relative to a moving observer
is
′
v = vsound ± vobserver
and the observed frequency is
The mass of the oil displaced is
M 2 = ρ o d2 A
= (874 kg/m3 ) (0.3 m) (0.03 m2 )
= 7.866 kg .
Since the weight of the block Mblock g is equal
to the weight of the displaced liquid, we have
Mblock g = [M2 + M3 ] g
Mblock = M2 + M3
= (7.866 kg) + (6.918 kg)
= 14.784 kg .
Question 20, chap 17, sect 4.
part 1 of 1
10 points
An ambulance is traveling north at
60.1 m/s, approaching a car that is also traveling north at 34.3 m/s. The ambulance driver
hears his siren at a frequency of 926 cycles/s.
The velocity of sound is 343 m/s.
60.1 m/s
34.3 m/s
Ambulance
14
Car
′
v
f = .
λ
′
Note: The wavelength is specified in the
reference frame of the medium of propagation.
Sound waves always travel at a given speed
with respect to their medium of propagation.
Note: Both the ambulance and car drivers,
as well as any observers at rest (on the side
of the road, for example), will measure the
same wavelength for the sound from the siren,
because length measurements will not depend
on the velocity of the measurer.
However, as sources/observers move
through the medium at different velocities,
they see the sound waves move past them at
different velocities. As a result, the number
of wavefronts passing them in a given time interval (i.e., the frequency of the sound) must
change.
The relationship between observed frequency and observed wavelength is always
given by
′
λ f = vrel ,
What is the wavelength at the car driver’s
position for the sound from the ambulance’s
siren?
Correct answer: 0.305508.
Explanation:
Let : vcar
vamb
vsound
f
= 34.3 m/s ,
= 60.1 m/s ,
= 343 m/s , and
= 926 cycles/s .
By the Doppler Effect, the wavelength of
the sound created by a source with rest frequency f and speed vsource is
v
± vsource
λ = sound
.
f
where vrel is the relative speed of the sound
wave and the observer/source.
Therefore, the wavelength of the sound
emitted in front of the ambulance is
′
vsound − vamb
f
343 m/s − 60.1 m/s
=
926 cycles/s
λ =
= 0.305508 m .
The negative sign arises because the ambulance driver is traveling in the same direction
as these sound waves and therefore perceives
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
I1
them as being slower than sound waves emit= 10 log10
+ 10 log10 (4)
ted when the ambulance is at rest. This reI0
sults in a smaller wavelength; intuitively, the
= β1 + 10 log 10(4)
wavefronts are compressed together by the
= 20 dB + 10 log10 (4)
motion of the siren.
= 26.0206 dB .
Question 21, chap 17, sect 2.
part 1 of 1
10 points
The sound of a man shouting at the top
of his lungs from a rather large distance away
from your ear has loudness of only 20 decibels.
What would be the decibel level of four men
shouting at the top of their (equally powerful)
lungs from the same distance away from you
ear? Assume that there is no interference
from superposed waves.
Question 22, chap 16, sect 3.
part 1 of 1
10 points
The figure shows two wave pulses that are
approaching each other.
P
1. β4 = 5 dB
2. β4 = 160 dB
3. β4 = 10 dB
1.
5. β4 = 14 dB
6. β4 = 26 dB correct
2.
8. β4 = 80 dB
9. β4 = 6 dB
10. β4 = 40 dB
Explanation:
The decibel level β is a logarithmic measure
of sound intensity I:
I
β = 10 log10
.
I0
Since we have assumed that there is no interference from superposed waves, four men
shouting together produce a sound intensity
four times greater than that of just one man
(I4 = 4 I1 ), so
4 I1
β4 = 10 log10
I0
Q
Which of the following best shows the shape
of the resultant pulse when the centers of the
pulses, points P and Q, coincide?
4. β4 = 60 dB
7. β4 = 20 dB
15
3.
4.
correct
5.
Explanation:
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
Notice that the two pulses have the same
width and amplitude.
Choosing the point P (the same as point Q
when the two pulses coincide) as the origin,
the two pulses can be described as:
P :
Q:
y1 =
y2 =
(
A , −d ≤ x ≤ d
A , −d ≤ x < 0
−A
,
0<x<d
Using the principle of superposition, the resultant pulse is
2 A , −d ≤ x < 0
y = y1 + y2 =
.
0 , 0<x<d
P
16
On a standing wave, the distance of the nearest two nodes is
d=
λ
,
2
where λ is the wavelength. In this problem,
x = 0 is a node, which can be seen by adding
y1 and y2 .
So the smallest positive value of x corresponding to a node is
1
λ
2
1
= (0.909091 m)
2
= 0.454545 m .
x=
Question 24, chap 11, sect 3.
part 1 of 1
10 points
Q
A thin hoop of radius 3.4 m and mass 12 kg
is suspended from a pivot on the hoop itself
as shown in the figure.
P +Q
Question 23, chap 16, sect 4.
part 1 of 1
10 points
θ
A standing wave is a superposition of two
harmonic waves described by
y1 = A sin(k x + ω t)
y2 = A sin(k x − ω t) ,
and
where A = 7.93302 cm, k = 6.9115 m−1 and
ω = 15.0796 s−1 .
Determine the smallest positive value of x
corresponding to a node .
Correct answer: 0.454545.
Explanation:
λ=
2π
k
2π
6.9115 m−1
= 0.909091 m .
=
The acceleration of gravity is 9.8 m/s2 .
Find the period T of small oscillations of
this hoop around θ = 0.
Correct answer: 5.23385.
Explanation:
The hoop oscillates as a physical pendulum.
Its center of mass is in its geometric center,
at distance R = 3.4 m from the pivot. So
when it swings by a small angle θ away from
the equilibrium position (center directly below the pivot), there is a torque due to gravity
force,
τ = −mg × R sin θ
(1)
≈ mgr × θ
midterm 04 – JYOTHINDRAN, VISHNU – Due: May 2 2007, 11:00 pm
for small θ.
The hoop has moment of inertia I0 = mR2
around its center of mass. The momend of
inertia relative to the pivot follows via the
parallel axis theorem
I = I0 + mR2 = 2mR2 .
(2)
Consequently, the equation of motion for the
swinging hoop
I
d2 θ
=τ
dt2
(3)
becomes
d2 θ
2mR2 × 2 = −mgR × θ,
dt
Explanation:
Basic Concepts: From
r
m
T = 2π
,
k
we have
4 π2 m
T2
4 π 2 (0.417 kg)
=
(0.922 s)2
= 19.3657 N/m .
k=
The total energy E is,
(4)
E=
which simplifies to
d2 θ
g
=−
×θ
2
dt
2R
(5)
This is a harmonic oscillator equation with
ω2 =
g
.
2R
(6)
Translating this angular frequency into the
oscillation period, we have
T =
2π
ωs
= 2π
= 2π
s
2R
g
(7)
2(3.4 m)
9.8 m/s2
= 5.23385 s.
Note that this period does not depend on the
hoop’s mass m.
Question 25, chap 15, sect 3.
part 1 of 1
10 points
A(n) 0.417 kg mass is attached to a spring
and undergoes simple harmonic motion with
a period of 0.922 s . The total energy of the
system is 2.1 J .
Find the amplitude of the motion.
Correct answer: 0.465702.
1
k A2 ,
2
where A is the amplitude of motion. Thus
r
2E
A=
s k
=
2 × (2.1 J)
(19.3657 N/m)
= 0.465702 m .
17