Kamal A.A. 1000 Solved Problems in Classical Physics: An Exercise Book
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5.3 Solutions 227
Fig. 5.19
dt
d θ
=
r
2
h
Time taken for the object to move from P
1
to P
2
(Fig. 5.19) is given by
t =
dt =
r
2
d θ
h
=
4a
2
h
θ
0
d θ
(1 + cos θ)
2
=
a
2
h
θ
0
sec
4
θ
2
d θ =
2a
2
h
θ
0
1 + tan
2
θ
2
d
tan
1
2
θ
=
2a
2
h
tan
1
2
θ +
1
3
tan
3
1
2
θ
But h =
√
GM × semi - latus rectum =
√
2aG M
∴ t =
2a
3
GM
tan
1
2
θ +
1
3
tan
3
1
2
θ
5.47 Required time for traversing the arc PQT is obtained by the formula derived
in problem (5.46), Fig. 5.20
t
0
= 2t = 2
2a
3
GM
tan
1
2
θ +
1
3
tan
3
1
2
θ
(1)
228 5 Gravitation
Fig. 5.20
For parabola
r =
2a
1 + cos θ
(2)
or cos θ =
2a
r
− 1(3)
∴ tan
θ
2
=
1 − cos θ
1 + cos θ
=
R
a
− 1(4)
where we have put r = R, the radius of earth’s orbit. Using (4) in (1)
t
0
=
2
3
2
GM
(2a + R)
√
R − a (5)
t
0
is maximized by setting
dt
0
da
= 0. This gives
a =
R
2
(6)
Using (6) in (5) gives
t
0
(max) =
4
3
R
3
GM
=
2
3π
2π
R
3
GM
=
2
3π
T
where T = 2π
R
3
GM
= 1 year is the time period of the earth.
Thus t
0
(max) =
2
3π
years.
5.3 Solutions 229
5.48
1
p
2
=
1
r
2
+
1
r
4
dr
dθ
2
(1)
r = a sin nθ (2)
dr
dθ
2
= n
2
a
2
(1 − sin
2
nθ) = n
2
a
2
1 −
r
2
a
2
(3)
Using (3) in (1)
1
p
2
=
n
2
a
2
r
4
+
1 − n
2
r
2
Differentiating
−
2
p
3
dp
dr
=−
4n
2
a
2
r
5
−
2(1 − n
2
)
r
3
or
1
p
3
dp
dr
=
2n
2
a
2
r
5
+
1 − n
2
r
3
Force per unit mass
f =−
h
2
p
3
dp
dr
=−h
2
2n
2
a
2
r
5
+
1 − n
2
r
3
5.49
1
p
2
=
1
r
2
+
1
r
4
dr
dθ
2
(1)
r = a(1 − cos θ)
dr
dθ
= a sin θ
dr
dθ
2
= a
2
sin
2
θ = a
2
1 −
1 −
r
a
2
= 2ra − r
2
(2)
Using (2) in (1)
1
P
2
=
2a
r
3
or p
2
=
r
3
2a
(3)
Force per unit mass
f =−
h
2
p
3
dp
dr
(4)
Differentiating (3)
230 5 Gravitation
2p
dp
dr
=
3r
2
2a
(5)
Using (5) in (4)
f =−
3
4
h
2
r
2
ap
4
=−
3ah
2
r
4
(6)
where we have used (3). Thus the force is proportional to the inverse fourth
power of distance.
5.50 If u =
1
r
, then at the apse
du
d θ
= 0
or −
1
r
2
dr
d θ
= 0
∴ −
1
r
2
a sin θ = 0
from which either sin θ = 0orr is infinite, the latter case being inadmissible
so long the particle is moving along the cardioid.
Thus θ = πor 0
When θ = π, r = 2a
and Q =
3ah
2
r
4
=
3ah
2
16a
4
=
3h
2
16a
3
Also p
2
=
r
2
2a
=
8a
3
2a
= 4a
2
and v
2
=
h
2
p
2
=
h
2
4a
2
Thus 4aQ=
3h
2
4a
2
= 3v
2
When θ = 0, r = 0 and p = 0 and the particle is moving with infinite velocity
along the axis of the cardioid and continues to move in a straight line.
5.51 Let the force f =−
k
r
3
=−ku
3
where u =
1
r
d
2
u
d θ
2
+ u =−
f
h
2
u
2
=
ku
h
2
5.3 Solutions 231
Case (i):
k
h
2
> 1
Let
k
h
2
− 1 = n
2
d
2
u
d θ
2
− n
2
u = 0
which has the solution u = Ae
nθ
+ Be
−nθ
, where the constants A and B
depend on the initial conditions of projection. If these are such that either A
or B is zero then the path is an equiangular spiral
Case (ii):
k
h
2
= 1, the equation becomes
d
2
u
d θ
2
= 0, whose solution is v =
Aθ + B, a curve known as the reciprocal spiral curve.
Case (iii):
k
h
2
< 1. Let 1 −
k
h
2
= n
2
, the equation becomes
d
2
u
d θ
2
+ n
2
u = 0
whose solution is u = A cos nθ + B sin nθ , a curve with infinite branches.
5.52
1
p
2
=
1
r
2
+
1
r
4
dr
d θ
2
(1)
r
2
= a
2
cos
2
θ (2)
r
dr
d θ
=−a
2
sin
2
θ (3)
∴
1
r
4
dr
d θ
2
=
a
4
r
6
sin
2
2θ =
a
4
r
6
1 −
r
4
a
4
=
a
4
r
6
−
1
r
2
(4)
From (1) and (4)
1
p
2
=
a
4
r
6
or p =
r
3
a
2
(5)
∴
dp
dr
=
3r
2
a
2
(6)
f =−
h
2
p
3
dp
dr
=−
3h
2
a
4
r
7
where we have used (5) and (6).
5.53 The polar equation of a circle with the origin on the circumference is r =
2a cos θ where a is the radius of the circle, Fig. 5.21.
232 5 Gravitation
Fig. 5.21
1
p
2
=
1
r
2
+
1
r
4
dr
dθ
2
(1)
r = 2a cos θ (2)
dr
d θ
=−2a sin θ (3)
∴
dr
dθ
2
= 4a
2
(1 − cos
2
θ) = 4a
2
−r
2
(4)
∴
1
r
4
dr
dθ
2
=
4a
2
r
4
−
1
r
2
(5)
Using (5) in (1) and simplifying
p =
r
2
2a
(6)
dp
dr
=
r
a
(7)
f =−
h
2
p
3
dp
dr
=−
h
2
a
4
r
5
5.54 Initially the earth’s orbit is circular and its kinetic energy would be equal to
the modulus of potential energy
1
2
mv
2
0
=
1
2
mGM
r
(1)
Suddenly, sun’s mass becomes half and the earth is placed with a new quan-
tity of potential energy, its instantaneous value of kinetic energy remaining
unaltered.
5.3 Solutions 233
New total energy = new potential energy + kinetic energy
=−G
M
2
m
r
+
1
2
mv
2
0
=−
GMm
2r
+
1
2
GMm
r
= 0
As the total energy E = 0 the earth’s orbit becomes parabolic.
Chapter 6
Oscillations
Abstract Chapter 6 deals with simple harmonic motion and its application to var-
ious problems, physical pendulums, coupled systems of masses and springs, the
normal coordinates and damped vibrations.
6.1 Basic Concepts and Formulae
Simple Harmonic Motion (SHM)
In SHM the restoring force (F) is proportional to the displacement but is oppositely
directed.
F =−kx (6.1)
where k is a constant, known as force constant or spring constant. The negative sign
in (6.1) implies that the force is opposite to the displacement.
When the mass is released, the force produces acceleration a given by
a = F/m =−k/m =−ω
2
x (6.2)
where ω
2
= k/m (6.3)
and ω = 2π f (6.4)
is the angular frequency.
Differential equation for SHM:
d
2
x
dt
2
+ ω
2
x = 0 (6.5)
Most general solution for (6.5)is
x = A sin(ωt + ε) (6.6)
235
236 6 Oscillations
where A is the amplitude, (ωt + ε) is called the phase and ε is called the phase
difference.
The velocity v is given by
v =±ω
A
2
− x
2
(6.7)
The acceleration is given by
a =−ω
2
x (6.8)
The frequency of oscillation is given by
f =
ω
2π
=
1
2π
k
m
(6.9)
where m is the mass of the particle.
The time period is given by
T =
1
f
= 2π
m
k
(6.10)
Total energy (E) of the oscillator:
E =
1
/
2
mA
2
ω
2
(6.11)
K
av
= U
av
=
1
/
4
mA
2
ω
2
(6.12)
Loaded spring:
T = 2π
M +
m
3
k
(6.13)
where M is the load and m is the mass of the spring.
If v
1
and v
2
are the velocities of a particle at x
1
and x
2
, respectively, then
T = 2π
x
2
2
− x
2
1
v
2
1
− v
2
2
(6.14)
A =
v
2
1
x
2
2
− v
2
2
x
2
1
v
2
1
− v
2
2
(6.15)
6.1 Basic Concepts and Formulae 237
Pendulums
Simple Pendulum (Small Amplitudes)
T = 2π
L
g
(6.16)
T is independent of the mass of the bob. It is also independent of the amplitude
for small amplitudes.
Seconds pendulum is a simple pendulum whose time period is 2 s.
Simple Pendulum (Large Amplitude)
For large amplitude θ
0
, the time period of a simple pendulum is given by
T = 2π
L
g
1 +
1
2
2
sin
2
θ
0
2
+
1.3
2.4
2
sin
4
θ
0
2
+
1.3.5
2.4.6
2
sin
6
θ
0
2
!
(6.17)
where we have dropped higher order terms.
Simple pendulum on an elevator/trolley moving with acceleration a. Time period
of the stationary pendulum is T and that of moving pendulum T
.
(a) Elevator has upward acceleration a
T
= T
g
g + a
(6.18)
(b) Elevator has downward acceleration a
T
= T
g
g − a
(6.19)
(c) Elevator has constant velocity, i.e. a = 0
T
= T (6.20)
(d) Elevator falls freely or is kept in a satellite, a = g
T
=∞ (6.21)
The bob does not oscillate at all but assumes a fixed position.