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Physics 1A PHYS1121 2008-T 2
Questions and Answers
Total Marks: 60
Question 1: 12 marks
- (a) Define the average coefficient of linear expansion for a material in terms of its fractional
change in length and the change in temperature. If your definition includes an equation, define all
terms used in it.
(b) A circular disk has a concentric circular hole cut out of its middle, as in the diagram.
Explain, giving your reasoning, what happens to both the disk and the hole as it is heated.
Suppose that r 0
=5.0cm, r 1
=10cm and the coefficient of linear expansion is 2.0x
°C
the change in the area of the material that makes up the disk for a 30°C increase in its temperature?
(c)
Water at temperature 25°C flows from a tap T into a heated container C. The container has a
heating element (a resistor R) which is supplied with electrical power, P, that may be varied.
The rate of flow of water F = 0.030 litres per minute. The electrical power is sufficient that the
water in the container is boiling. What is the minimum power that must be supplied in the steady
state so that the amount of liquid water in the container neither increases nor decreases with time?
You may neglect other losses of heat, such as conduction and radiation from the container to the air.
For water, c=4,200 J kg
K
, L
vap
=2.30x
6
J kg
, ρ=1,000 kg m
Solution
(a) The coefficient of linear expansion, α, for a material is defined as the fractional change in
length, divided by the change in temperature.
i.e.
α =
Δ L / L
i
Δ T
where ΔL is the change in length, L f
- L
i
, ΔT is the change in temperature, T f
- T
i
and
i, f refer to the initial and final values for the relevant quantities.
(b) Both the disk and hole expand in radius (and in area) as the disk is heated. All elements
expand by the same fraction, so that both the radius of a circle, and its circumference, will expand
by the same fraction. The areas will also expand. Thus both the hole and the disk expands.
r
o
r
1
We have ΔA = 2α A i
ΔT for area expansion (or derive directly from square of linear expansion). i.e.
twice the rate of linear expansion.
With A disk
= π (r 1
2
2
) for the disk and A hole
= π r 0
2
Thus, ΔA disk
= 2 x 2.0 10
x π x [
2
2
] x 30 = 0.2827 cm
2
for disk
= 0.28 cm
2
for disk to 2SF
And ΔA hole
= 2 x 2.0 10
x π x 5.
2
x 30 = 0.09425 cm
2
for hole
= 0.094 cm
2
for hole to 2SF
(c) We must have, in the steady state,
Energy supplied = Heat to raise water to boiling + Heat to boil the water
i.e. P t = m L + m c ΔT
where m is the mass of the water in the container, L and c are the latent and specific heats,
respectively, and ΔT is the temperature rise, in time t.
The flow rate, F, given by m = ρ F t where ρ is the density.
Thus P t = ρ F t [L + c ΔT]
or P = ρ F [L + c ΔT]
i.e. P = 1000 x 0.030 x 10
/60 x [2.3 10
6
= 1307.5 W = 1,300W to 2SF.
When letting air rapidly out of a bicycle tyre, the change is adiabatic, so Q=0. The work done is
negative since the gas expands (the gas does work). Thus, there is a decrease in internal energy,
applying the 1
st
Law. Hence a decrease in temperature, since this depends on the internal energy.
When allowing the pan of hot water to cool, there is no work done as the water is simply cooling. It
loses energy as it does so. Thus Q<0. So, from the 1
st
Law, ΔE
int
is < 0, and hence so is ΔT.
(c)
There is rapid compression from position 1 to position 2 along (a)
There is a decrease in pressure at constant volume along (b)
There is a slow increase in volume along (c).
Path (a) is adiabatic, as the process is fast and there is no time for heat to flow. So Q=0, but P and T
both rise.
Path (b) is isovolumetric (or isochoric) and heat flows out of the gas as the temperature falls to 0°C.
Path (c) is isothermal because the expansion is slow enough to remain in thermal equilibrium with
the bath. i.e. it takes place at constant temperature.
When we return to the start, the gas is at the same state, so at the same internal energy.
Thus ΔE int
= 0, so from the 1
st
Law Q + W = 0.
Thus W = – Q = heat to melt the ice
= – (-0.2 x 3.33 10
5
J/kg) = 6.66 x 10
4
J/kg.
Since Q is < 0, as heat is lost from the gas to the ice-water, so that the work done on the gas, W, is
positive.
(a)
(c)
(b)
P
V
Start
Question 3 11 Marks
- (a) Show that x = A cos( ω t+ φ ) describes the displacement of a system where the
restoring force is proportional to the distance from an equilibrium position, x.
(b) What do the symbols A, ω and φ correspond to?
(c) Hence show that, for a stretched spring of spring constant k , with block of mass m attached
at the free end, that the period of oscillation, T , of the block is given by
T = 2 π
m
k
(d) For this system, give expressions for the kinetic energy and potential energy of the block at
position x , in terms of A, m, k and φ.
(e) What also is the total energy of the system at this position? Simplify your expression and
comment on this value.
(f) Suppose that k =5.00 N/m and the block has mass m =0.200kg, and is oscillating with an
amplitude of 10.0cm. If initially the velocity is 0.100 m/s in the negative x - direction, determine the
equation describing the distance x from equilibrium as a function of time. What is the maximum
speed and acceleration experienced by the block?
Solution
(a) Let that x = A cos( ω t+ φ )
Then
v =
x = − ω A sin( ω t + φ)
And
a =
x = − ω
2
A cos( ω t + φ) = − ω
2
x
So that
F
x
= m
x = − m ω
2
x , as required for the restoring force being proportional to the
distance from the equilibrium.
(b) A is the amplitude of the oscillation
ω is the angular frequency
φ is the phase angle (i.e. the phase when t=0)
(c) For SHM T=2π/ω.
For a spring, where
F
x
= m x ′′= − kx with ω
2
=k/m.
Thus
T = 2 π
m
k
(d) KE at position x is
1 2 mv
2
= 1 2 m − ω A sin( ω t + φ)
[ ]
2
= 1 2 m ω
2
A
2
sin
2
( ω t + φ)
PE at position x is
1 2 kx
2
= 1 2 kA
2
cos
2
( ω t + φ)
Thus, since k=mω
2
, then KE = 1/2kA
2
sin
2
(ωt+φ)
and PE = 1/2kA
2
cos
2
(ωt+φ)
i.e. these are independent of the mass, and only involve A, k and φ.
(e) The total energy E = KE + PE = 1/2k A
2
since sin
2
2
This is fixed, as it must be for conservation of energy.
Question 4. 14 marks
(a) Two travelling waves, moving in the +x and – x directions, are described by the equations
y=C sin(kx- ω t) and y=C sin(kx+ ω t) , respectively. What do the symbols k and ω represent, and how
are they related to the wave speed along the x - direction?
(b) Derive an expression for the wave resulting from the linear superposition of these two
travelling waves. Interpret this equation quantitatively with the aid of a labelled sketch.
(c) For a thin cylindrical pipe of length L , open at both ends, and sound speed c , show that the
frequency of the resonant modes of oscillation is given by
f =
nc
2 L
, where n is an integer, ≥ 1.
Sketch the first three resonant modes.
(d) Suppose that a small loudspeaker with frequency adjustable from 1 to 2 kHz is placed near
to, but not touching, one end of the pipe. The pipe has length 0.60m. At what frequencies will
resonance occur? Take the speed of sound to be 330 m/s.
(e) Describe how taking into account end effects will alter the resulting resonant frequencies
from part (d).
Solution
(a) k is the wavenumber = 2π/λ, where λ is the wavelength
ω is the angular frequency = 2 π/T, where T is the period
v = ω /k is the wave speed, for the disturbance travelling in either the +x or – x direction.
(b)
Let y 1
=C sin(kx- ω t) and y
2
=C sin(k+ ω t).
Then the sum, y=y 1
+y 2
=C[sin(kx- ω t) +sin(kx+ ω t)]
Making use of sinA + sinB = 2 sin(A+B)/2 cos(A-B)/2 then,
With A= kx- ω t and B= kx+ ω t , we have
Y=2 sin(kx) cos(- ω t) = 2 sin(kx) cos( ω t).
Nodes are positions with no oscillation, so sin(kx)=0, hence kx=nπ, with n=0, 1, 2, …
Antinodes are positions with maximum oscillation, sin(kx)=+/-1, hence kx=(n+1/2) π with n=0, 1,
Each element of the string oscillates in SHM given by cos( ω t) within an envelope of amplitude
2Csin(kx ).
(c)
2C
C
π /k
π /k
π /k
2 π /k
Node
Antinode
π /2k
3 π /2k
2Csin(kx)
Question 5. 8 Marks
(a) Suppose that a person (the observer) is moving at speed v O
directly towards a stationary source
of sound which emits waves of frequency f at wavespeed c in still air. Show that the frequency
experienced by the person, f ′, is given by
f ′= f
c + v
O
c
(b) A person walks at 5 m/s directly towards a stationary siren emitting sound at frequency
1,500 Hz in still air. What frequency will the person hear? The speed of sound may be taken as
330 m/s.
(c) Suppose now that the siren is also moving directly away from that person at a speed of 20
m/s. What frequency will the person now experience?
Solution
(a)
Speed of sound c = f λ
Observer travels towards source at speed V obs.
Speed of waves relative to observer is now c+V
obs
Wavelength is unchanged. Thus observed frequency, f ′ given by
f ′=
c + V
obs
λ
= f
c + V
obs
c
(b) We have V
Obs
= 5 m/s, f=1500 Hz and c=330 m/s
So
f ′= f
c + V
obs
c
Hz
Thus f ′=1523 Hz
(c) The general Doppler effect equation is
f ′= f
c + V
Obs
c − V
Source
where V
Obs
and V
Source
are directed
towards each other.
In this case, V Obs
= 5 m/s and V Source
= - 20 m/s as the source is moving away from the observer.
Thus
f = 1500
= 1436 Hz
X
V
obs
c
f
