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PHYS1131 Higher Physics 1A
2009 - T1 Test 2 Solutions
Thermal Physics and Waves
3 (a) Take three from:
(i) large number of molecules, average separation large compared to their
dimensions – i.e., point-like
(ii) Don’t interact except during collisions
(iii) Elastic collisions with walls
(iv) Obey Newton’s 2
nd Law; but move randomly (v) Pure substance (i.e. all particles the same)
(b) Explain elastic collisions or show diagram as below:
Speed vx, momentum mvx
Change in momentum on collision of a particle with the wall is
p = mvx – (-mvx) = 2mvx
Average time between collision of particle with the same wall τ =
2 d
vx
Thus Force exerted, F = rate of change of momentum = N
Δ p
τ
i.e.
F =
N 2 mvx
2 d vx
= N
mvx
2
d
(c) Pressure P =
Force
Area
F
d
2 =^
Nmvx
2
d
3 =^
Nmvx
2
V
Where V = d
3 = volume of box
But
v
2 = vx
2
2
2
and from isotropy
vx
2 = vy
2 = vz
2 i.e.
vx
2
v
2
So that P =
N
3 V
mv
2
d
Vx
Vx
m
(d) Equation of state for an ideal gas is given by PV = nRT
Where P is the pressure
V is the volume
n is the number moles of the gas
R is the gas constant
T is the absolute temperature
Also OK to use PV = N k T where N is the number of particles, k Boltzmann’s constant
(e) Kinetic energy K = N x KE per particle = N x ½ mv
2
So
K =
N 3 PV
2 N
= 1.5 PV = 1.5 nRT from the ideal gas law
(f) T = 20 C = 293 K
MO 2 = 2 x 16 mp
From part (e) we have K = 3/2 n RT = N ½ mv
2
i.e. v
2
But N = n NA where NA is Avogadro’s Number
V =
3 × 8.314 × 293
32 × 1.
− 27 × 6.022 × 10
23
m
2 s
v
2 = 2.267 × 10
5 m
2 /s
2
So that v = 476.1 m/s = 480 m/s to 2SF
Q5 (a). fsource = 262.0 Hz, vsource = 3 m/s, vsound wave = 340 m/s
For a moving musician and stationary observer we have:
f observer = f source
v
v − vsource
= 264.33 Hz
using the Doppler formula with v Observer= 0 m/s.
So f Observer = 264.33 Hz. But f Source = 262.0 Hz i.e. unchanged.
Thus, the difference in frequencies = 264.3 – 262.
= 2.3 Hz to 2SF
(b) We now have v observer = 30 m/s moving towards the musician
Thus,
f observer = f source
v + v Observer
v − vsource
from the Doppler formula
For moving observer, stationary source fObs,1 = 262.(340+30)/
= 285.12 Hz
For moving observer, moving source fObs,2 = 262.(340+30)/(340-3)
= 287.66 Hz
Thus, the difference in frequencies Δ f = 287.66 – 285.
= 2.54 Hz
= 2.5 Hz to 25F
Observer
Vsource
V=340 m/s
Vobs
Vsource
V
(c) Antinode occurs where the string is plucked.
Node occurs at both fixed ends
Guitar plucked in middle at L/2 Guitar plucked at L/
For guitar string plucked in middle
λ
2
= L ⇒ λ = 2 L
For guitar string plucked at L/6:
λ
4
L
⇒ λ =
2 L
Since c=f λ we have f=c/2L in L/2 case
f=3c/2L in L/6 case
Hence, frequency increases by a factor [ 3c/2L ] / [ c/2L ] = 3 times
(d) Organ pipe open at both ends. Thus antinodes at the ends.
End effects are ignored.
First resonant mode looks like:
Node
L = λ/2 λ/4 = L/
Antinode
Antinode
L = λ/
Antinode
- (a) SHM: E=Total Energy = PE + KE
≡ Max value of PE
≡ Max value of KE
Suppose PE = ½ kx
2
Then
PE max = 1 2
kx 0
2 = E when x 0 is the maximum extension.
So when x = x 0 /2 then
PE = 1
k (
x 0
2
2 = 1 4
PE max = E^ 4
Thus, KE when x = xo is E-E/4 = 3E/4.
(b)
(i) Block is in the free fall when only under the influence of gravity.
i.e., for the first 11.0m of the fall, when the cord does not retard its motion
So using
" s = ut +
at
2 " for this portion of the fall we have:
t =
2 s
a
since u = 0.
∴ t =
2 × 11.
s = 1.498s = 1.5s to 1d.p.
h = 36 m
Free-fall L=11 m
SHM
M
(ii) Conserving energy, and neglecting air resistance,
PE lost by falling block = PE in cord + KE of falling block.
At maximum distance, h , block no longer moving, so then has no KE,
Thus,
mgh = 1 2
k ( h − L )
2
So that
k =
2 mgh
( h − L )
2 =^
2 N/m
Hence k= 73.38 N/m = 73 N/m to 2SF.
(iii) Equation of motion is given by N2L; i.e. Mass x Acceleration = Force = Restoring Force Spring + Weight of Block
i.e.
m ˙ x ˙ = − kx + mg
But
mg = kx 0 for the equilibrium extension of the spring.
∴ m ˙ x ˙ = − k ( x − x 0 )
∴ m ˙ x ˙ ' = − kx '
where x’=x-x 0 is the extension from equilibrium
This is the equation for SHM hence the motion is SMH.
(iv) SHM with
ω
2
= k / m = (73.4 /65.0)
∴ ω=1.06 rad/s = 1.1 rad/s to 1 d.p.
(v) Equilibrium extension of the spring given by:
x 0 =
mg
k
m = 8.68m = 8.7m to 2SF
