Since the lock-in amplifier measures the ac signal of the same frequency with its reference signal, the frequency of the piezoelectric tube oscillation, the frequency of the... cantilev[r]
(1)Theoretical Question 3: Scanning Probe Microscope
1 Answers
(a)
2 2 2
0 )
(ω ω b ω m
F A
+ −
= and
) (
tan 2 2
0
ω ω
ω φ
− =
m b
At ω =ω0,
0 ω
b F
A=
and
π
φ =
(b) A non-vanishing dc component exists only when ω =ωi In this case the amplitude of the dc signal will be Vi VR cosφi
2
0
0
(c)
0
0
2 bω V c
c R
at the resonance frequency ω0
(d) ∆m=1.7×10−18 kg
(e)
2 /
0
'
− =
ω ω
ω
m c
(f)
3 /
0
0
∆ =
ω ω
m qQ k
d e
=
0
(2)2 Solutions (a) [1.5 points]
Substituting z(t)= Asin(ωt−φ) in the equation m z F t dt
dz b dt
z d
m ω2 0sinω
0
2
= +
+
yields,
t A F t
m t
b t
mω sin(ω φ) ωcos(ω φ) ω 2sin(ω φ) 0sinω
2 − + − + − =
− (a1)
Collecting terms proportional to sinωt and cosωt, one obtains
{ ( )sin cos }cos
sin sin
cos )
( 2
0
2
0 + − − + =
− + − t m b t
A F b
m ω ω φ ω φ ω ω ω φ ω φ ω (a2)
Zeroing the each curly square bracket produces ) (
tan 2 2
0 ω ω
ω φ
− =
m b
, (a3)
2 2 2
0 )
(ω ω b ω m
F A
+ −
= (a4)
At ω =ω0,
0 ω
b F
A= and
2
π
φ = (a5)
(b) [1 point]
The multiplied signal is
}] ) cos{(
} ) [cos{(
2
) sin( )
sin( 0
0
i i
i i
R i
R i i i
t t
V V
t V
t V
φ ω ω φ
ω ω
ω φ
ω
− + −
− − =
−
(b1) A non-vanishing dc component exists only when ω=ωi In this case the amplitude of the dc signal will be
i R i V
V cosφ
2
0
0 (b2)
(c) [1.5 points]
(3)cantilever, and the frequency of the photodiode detector should be same The magnitude of the input signal at the resonance is
0 0
0 ω bω
V c c b F c V R
i = = (c1)
Then, since the phase of the input signal is 2 + =
−π π at the resonance, φi =0 and the lock-in amplifier signal is
0 2 0 cos ω b V c c V V R R
i = (c2)
(d) [2 points]
The original resonance frequency
m k
=
0
ω is shifted to
− ∆ = − ∆ ≅ + ∆ = ∆ + − m m m m m k m m m k m m k 1 1
1 0
1
ω (d1)
Thus m m ∆ − =
∆ 0 0
2 1ω
ω (d2)
Near the resonance, by substituting φ→π +∆φ
2 and ω0 →ω0 +∆ω0 in Eq (a3), the change of the phase due to the small change of ω0 (not the change of ω) is
0 tan tan ω φ φ π ∆ = ∆ − = +∆ m b
(d3)
Therefore,
b
m 0
2
tan φ ω
φ ≈ ∆ =− ∆
∆ (d4)
From Eqs (d2) and (d4),
18 18 12 10 10 1800 10 10
10 ⋅ − = − = × −
= ∆ = ∆ φ π π ω b
m kg (d5)
(e) [1.5 points]
In the presence of interaction, the equation of motion near the new equilibrium position
(4)t F z c z m dt dz b dt
z d
m ω2 3 0sinω
0
2
= − +
+ (e1)
where we used f(h)≈ f(h0)+c3z with z=h−h0 being the displacement from the new equilibrium position h0 Note that the constant term f(h0) is cancelled at the new equilibrium position
Thus the original resonance frequency
m k
=
0
ω will be shifted to
2
3
0
'
ω ω
ω ω
m c m
c m m
c
k− = − = −
= (e3)
Hence the resonance frequency shift is given by
− −
=
∆ 2
0
0 ω ω
ω
m c
(e4)
(f) [2.5 points]
The maximum shift occurs when the cantilever is on top of the charge, where the interacting force is given by
2 )
(
h qQ k h
f = e (f1)
From this,
3
3
0 d
qQ k dh
df
c e
d h
− = =
=
(f2) Since ∆ω0 <<ω0, we can approximate Eq (e4) as
0 2 ω
ω
m c
− ≈
∆ (f3)
From Eqs (f2) and (f3), we have
3 0
0
0
2
d m
qQ k d qQ k
mω e e ω
ω =
− −
=
∆ (f4)
(5)8
/
0
0 4.1 10
− × =
∆ =
ω ω
m qQ k
d e m = 41 nm (f5)
(6)3 Mark Distribution
No Total Pt
Partial
Pt Contents
0.7 Equations for A and φ (substitution and manipulation) 0.4 Correct answers forA and φ
(a) 1.5
0.4 A and φ at ω0
0.4 Equation for the multiplied signal
0.3 Condition for the non-vanishing dc output (b) 1.0
0.3 Correct answer for the dc output 0.6 Relation between Vi and VR
0.4 Condition for the maximum dc output (c) 1.5
0.5 Correct answer for the magnitude of dc output 0.5 Relation between ∆m and ∆ω0
1.0 Relations between ∆ω0 (or ∆m) and ∆φ (d) 2.0
0.5 Correct answer (Partial credit of 0.2 for the wrong sign.)
1.0 Modification of the equation with f(h) and use of a proper approximation for the equation
(e) 1.5
0.5 Correct answer
0.5 Use of a correct formula of Coulomb force 0.3 Evaluation of c3
0.6 Use of the result in (e) for either ∆ω0 or ω'20−ω02 0.6 Expression for d0
(f) 2.5
0.5 Correct answer