Example 16: Show that the free oscillations of a galvanometer needs as affected by the viscosity of the surrounding air which varies directly as the angular velocity of the needle are determined by the equation dt dt where k is the coefficient of viscosity and 0 is the angular deflection of the needle at time t. Obtain e in terms of t and discuss cases that can arise. Ttangent PT) Solution: The geometry of the problem (Fig. 4.10) is identical to one already explained in article 10.5. Here, simply one additional force viz resistance due to air acts in the direction opposite to that of m . The equation of motion along the tangent PT at P is g sin e mg df Fig. 10.10 ge- e) dt ((e) -mg sine -a (for small e, sine = 0) or d fe de Le =0 or m dt e .+ ue =0 or dt df dt where k=- is the coefficient of viscosity and u= Corresponding auxiliary equation being (D² + kD+µ)=0 implyîng D=±±F- 4u . Various cases for complementary solution are as follows: 2 Case 1: when F> 4. 8 = G c where i = F – 4u +Ce Case 2: When F = 4u., 8=(C, +Cf)e; case of equal roots Case 3: When F< 4u e= Ge ? +Ge (p= Ju –F) :D= e=e (G,cos pt + G; sin pt). case of complex roots with G= rcose) C=rsino] Let so that r=JG + G and o= tan- Therefore, 8= (rcosocos pt + rsinosin pt) =reT cos(pt -0) However, the motion of the pendulum will be simple harmonic only in case 3, where < 4µ.
Example 16: Show that the free oscillations of a galvanometer needs as affected by the viscosity of the surrounding air which varies directly as the angular velocity of the needle are determined by the equation dt dt where k is the coefficient of viscosity and 0 is the angular deflection of the needle at time t. Obtain e in terms of t and discuss cases that can arise. Ttangent PT) Solution: The geometry of the problem (Fig. 4.10) is identical to one already explained in article 10.5. Here, simply one additional force viz resistance due to air acts in the direction opposite to that of m . The equation of motion along the tangent PT at P is g sin e mg df Fig. 10.10 ge- e) dt ((e) -mg sine -a (for small e, sine = 0) or d fe de Le =0 or m dt e .+ ue =0 or dt df dt where k=- is the coefficient of viscosity and u= Corresponding auxiliary equation being (D² + kD+µ)=0 implyîng D=±±F- 4u . Various cases for complementary solution are as follows: 2 Case 1: when F> 4. 8 = G c where i = F – 4u +Ce Case 2: When F = 4u., 8=(C, +Cf)e; case of equal roots Case 3: When F< 4u e= Ge ? +Ge (p= Ju –F) :D= e=e (G,cos pt + G; sin pt). case of complex roots with G= rcose) C=rsino] Let so that r=JG + G and o= tan- Therefore, 8= (rcosocos pt + rsinosin pt) =reT cos(pt -0) However, the motion of the pendulum will be simple harmonic only in case 3, where < 4µ.
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![Physical Application of Linear Differential Equations
651
Example 16: Show that the free oscillations of a galvanometer needs as affected by the
viscosity of the surrounding air which varies directly as the angular velocity of the needle
are determined by the equation
d'e
dt
where k is the coefficient of viscosity and 0 is the angular o
deflection of the needle at time t. Obtain e in terms of t and
discuss cases that can arise.
T(tangent PT)
Solution: The geometry of the problem (Fig. 4.10) is
identical to one already explained in article 10.5. Here,
simply one additional force viz resistance due to air acts in
the direction opposite to that of ms
mg sin e
mg
dt
The equation of motion along the tangent PT at P is
Fig. 10.10
d's
de
F ((e)
d(1e)
=-ng sine -a
ds
or
=-mge -a
dt
(for small 0, sine = 0)
dt
df
fe
fe
de
a de
+ 소8=D0
m dt
+ k-
+ ue = 0
or
or
dt?
dt
dt
where k=4 is the coefficient of viscosity and u=
Corresponding auxiliary equation being
-kt Jk? - 4u
(D + kD +u)=0 implying D=-
.. Various cases for complementary solution are as follows:
Case 1: when k > 4µ, e = Ge+C,e, where = - 4µ
kt
Case 2: When k = 4µ, 0 = (C, +C,t)e 7; case of equal roots
(++iple
+Ce 2
-k tip
(p= Jau - )
Case 3: When k< 4µ, 0= Ce ?
: D=
%3D
4u – R
e= e? (G,cos pt + C sin pt). case of complex roots with p=
G = rcoso]
Let C =rsino
so that
r= G + G and o = tan-
-kt
Therefore, e = eT (rcosocos pt + rsinosin pt)
= re? cos (pt -0)
However, the motion of the pendulum will be simple harmonic only in case 3, where < 4u.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9bdbc044-9372-40ac-ad69-d728d9c89027%2F8f3388cf-56d6-4d6a-b180-3275c82b9964%2Fsd779tx_processed.png&w=3840&q=75)
Transcribed Image Text:Physical Application of Linear Differential Equations
651
Example 16: Show that the free oscillations of a galvanometer needs as affected by the
viscosity of the surrounding air which varies directly as the angular velocity of the needle
are determined by the equation
d'e
dt
where k is the coefficient of viscosity and 0 is the angular o
deflection of the needle at time t. Obtain e in terms of t and
discuss cases that can arise.
T(tangent PT)
Solution: The geometry of the problem (Fig. 4.10) is
identical to one already explained in article 10.5. Here,
simply one additional force viz resistance due to air acts in
the direction opposite to that of ms
mg sin e
mg
dt
The equation of motion along the tangent PT at P is
Fig. 10.10
d's
de
F ((e)
d(1e)
=-ng sine -a
ds
or
=-mge -a
dt
(for small 0, sine = 0)
dt
df
fe
fe
de
a de
+ 소8=D0
m dt
+ k-
+ ue = 0
or
or
dt?
dt
dt
where k=4 is the coefficient of viscosity and u=
Corresponding auxiliary equation being
-kt Jk? - 4u
(D + kD +u)=0 implying D=-
.. Various cases for complementary solution are as follows:
Case 1: when k > 4µ, e = Ge+C,e, where = - 4µ
kt
Case 2: When k = 4µ, 0 = (C, +C,t)e 7; case of equal roots
(++iple
+Ce 2
-k tip
(p= Jau - )
Case 3: When k< 4µ, 0= Ce ?
: D=
%3D
4u – R
e= e? (G,cos pt + C sin pt). case of complex roots with p=
G = rcoso]
Let C =rsino
so that
r= G + G and o = tan-
-kt
Therefore, e = eT (rcosocos pt + rsinosin pt)
= re? cos (pt -0)
However, the motion of the pendulum will be simple harmonic only in case 3, where < 4u.
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