2 = mg . If is changed to – , then the restoring force is d d x O
Hints: 01 2 = mg . If is changed to – , then the restoring force is d d x O
MM PP
µ iiL
MM
g 2 π d d PP Q
µ 01 iiL 2 µ 01 iiL 2 mgx
+ mg ≈ –
64. You have a 2.0 mH inductor and wish to make an LC circuit whose resonant fre- quency can be tuned across the AM radio band (550 kHz to 1600 kHz). What range of capacitance should your variable capacitor cover?
SIMPLE HARMONIC MOTION 57
65. An object of mass 0.2 kg executes simple harmonic oscillations along the x-axis with
a frequency of (25/π) Hz. At the position x = 0.04 m, the object has kinetic energy
0.5 J and potential energy 0.4 J. Find the amplitude of oscillations. (I.I.T. 1994)
K.E. ( A – x ) O
MM N
x 2 Q PP
Hints: =
P.E.
66. T 1 is the time period of a simple pendulum. The point of suspension moves vertically upwards according to y = kt 2 where k = 1 m/s 2 . Now the time period is T 2 . Then T 2
(I.I.T. 2005)
6 [Hints : Upward acceleration of the point of suspension is a = 2k = 2 m/s 2 and in this
case the effective g is (10 + 2) m/s 2 ]
67. A simple pendulum has a time period T 1 when on the earth’s surface and T 2 when taken to a height R above the earth’s surface where R is the radius of the earth. Show
that the value of (T 2 /T 1 )
is 2.
GM O
Mm
GM
Hints : mg = G 2 , T 1 = 2 π l
R 2 2 = 2 π NMM l r 4 R 2 QPP
68. A particle executes simple harmonic motion between x = – A to x = + A. The time taken for it to go from 0 to A/2 is T 1 and to go from A/2 to A is T 2 . Show that T 2 /T 1 = 2.
π π Hints: x = sin A ω t , ω T 1 =
NM
6 2 QP
Superposition Principle and Coupled Oscillations
2.1 DEGREES OF FREEDOM
Number of independent coordinates required to specify the configuration of a system completely is known as degrees of freedom.
2.2 SUPERPOSITION PRINCIPLE
For a linear homogeneous differential equation, the sum of any two solutions is itself a solution.
Consider a linear homogeneous differential equation of degree n:
n − dy 1 d y
dy
n + a n − 1 n − 1 + .... + a 1 + ay 0 = 0 .
If y 1 and y 2 are two solutions of this equation then y 1 +y 2 is also a solution, which can
be proved by direct substitution.
2.3 SUPERPOSITION PRINCIPLE FOR LINEAR INHOMOGENEOUS EQUATION
Consider a driven harmonic oscillator
dx 2 m 2 =− kx + Ft ()
dt
where F(t) is the external force which is independent of x. Suppose that a driving force
F 1 (t) produces an oscillation x 1 (t) and another driving force F 2 (t) produces an oscillation x 2 (t) [when F 2 (t) is the only driving force]. When the total driving force is F 1 (t) + F 2 (t), the
corresponding oscillation is given by x(t) = x 1 (t) + x 2 (t).
2.4 SUPERPOSITION OF SIMPLE HARMONIC MOTIONS ALONG A STRAIGHT LINE
If a number of simple harmonic motions along the x-axis x i =a i sin (ω i t+φ i ), i = 1, 2, .., N
SUPERPOSITION PRINCIPLE AND COUPLED OSCILLATIONS 59
are superimposed on a particle simultaneously, the resultant motion is given by
X= ∑ x i = ∑ a i sin(ω i t+φ i ).
2.5 SUPERPOSITION OF TWO SIMPLE HARMONIC MOTIONS AT RIGHT ANGLES TO EACH OTHER
If two simple harmonic motions
x = a sin ω 1 t,
...(2.4) act on a particle simultaneously perpendicular to each other the particle describes a path
y = b sin(ω 2 t + φ)
known as Lissajous figure when ω 1 and ω 2 are in simple ratio. The equation of the path is obtained by eliminating t from these two equations. The position of the particle in the xy plane is given by
r r = xi $ + yj $
SOL SOL SOL SOL SOLVED PR VED PROBLEMS VED PR VED PR VED PR OBLEMS OBLEMS OBLEMS OBLEMS
1. Two simple harmonic motions of same angular frequency ω
x 1 =a 1 sin ωt, x 2 =a 2 sin (ωt + φ)
act on a particle along the x-axis simultaneously. Find the resultant motion.
Solution
The resultant displacement is X=x 1 + x 2 = sin ωt [a 1 +a 2 cos φ] + cos ωt [a 2 sin φ].
...(2.7) so that
R sinθ = a 2 sin φ
The resultant displacement is
...(2.10) which is also simple harmonic along the x-axis with the same angular frequency ω. The
X = R sin(ωt + θ)
amplitude R and the phase angle θ of the resultant motion are given by Eqns. (2.8) and (2.9) respectively.
Special Cases (i) φ = + 2 nπ, n = 0, 1, 2,.... or, the two SHMs x 1 and x 2 are in phase,
60 WAVES AND OSCILLATIONS
R=a 1 + a 2
(ii) φ = + (2n + 1)π, n = 0, 1, 2,...or, the two SHMs x 1 and x 2 are in opposite phase,
R=a 1 ~ a 2 .
In this case, the resultant amplitude is zero when a 1 =a 2 and one motion is destroyed by the other.
2. Find the resultant motion due to superposition of a large number of simple harmonic motions of same amplitude and same frequency along the x-axis but differing progressively in phase.
Solution
The simple harmonic motions are given by
x 1 = a sin ωt, x 2 = a sin(ωt + φ),
x 3 = a sin(ωt + 2φ),
§ x N = a sin[ωt + (N −1)φ].
The resultant displacement is
i = a sin ωt [1 + cos φ + cos 2φ +...+ sin (N –1) φ], + a cos ωt [0 + sin φ + sin 2φ +...+ sin (N –1) φ],
...(2.11) where
= R sin (ωt + θ)
R cos θ = a [1 + cos φ + cos 2φ +.... + cos (N – 1) φ], R sin θ = a [0 + sin φ + sin 2φ +.... + sin (N – 1) φ]
e i φ ( e ( N − 1 ) iφ φ 2iφ i(N – 1)φ − 1 Now, e +e + ...... + e
Equating the real and imaginary parts, we get
cos N φ / 2 sin ( N − 1 )/ φ 2
cos φ + cos 2φ +... + cos (N – 1) φ =
sin φ / 2 sin N φ / 2 sin( N − 1 ) φ / 2
sin φ + sin 2φ +...+ sin (N – 1)φ =
sin φ / 2
Thus, we write cos N φ / sin( 2 N − 1 )/ φ 2
1 + cos φ + cos 2φ + ...+ cos (N – 1) φ = 1 +
sin / φ 2
sin{ N − ( N − 1 )} / φ 2 cos N φ / sin( 2 N − 1 )/ φ 2
= sin / φ 2
sin N φ / cos ( 2 N − 1 )/ φ 2 = sin / φ 2
SUPERPOSITION PRINCIPLE AND COUPLED OSCILLATIONS 61
The resultant motion of Eqn. (2.11) is simple harmonic with amplitude and phase angle given by
sin( Nφ /) 2
...(2.13) When N is large and φ is small, we may write
and the phase difference between the first component vibration x 1 and Nth component vibration x N is nearly equal to 2θ.
The resultant amplitude may be obtained by the vector polygon method (Fig. 2.1). The polygon OABCD is drawn with each side of length a and making an angle φ with the neighbouring side. The resultant has the amplitude OD with the phase angle = ∠ DOA with respect to the first vibration.
Fig. 2.1
Special Cases A (i) We consider the special case when there is
superposition of a large number of vibrations x i of very small amplitude a but continuously increasing phase. The polygon will then become an arc of a circle and the chord joining the first and the last points of the arc will represent the amplitude of the resultant vibration (Fig. 2.2). When
the last component vibration is at A, the first and the last component vibration are in opposite phase and the amplitude of the resultant vibration = OA = diameter of the circle. When the last component vibration is at B, the first and the
last component vibrations are in phase, the polygon becomes
a complete circle and the amplitude of the resultant vibration
is zero.
Fig. 2.2
(ii) When the successive amplitudes of a large number of component vibrations decrease slowly and the phase angles increase continuously the polygen becomes a spiral converging asymptotically to the centre of the first semicircle.
62 WAVES AND OSCILLATIONS
3. The displacement y of a particle executing periodic motion is given by
2 F y = 4 cos 1 t I sin (1000 t). HG 2 KJ
show that this expression may be considered to be a result of the superposition of three independent harmonic motions.
(I.I.T. 1992)
Solution
I HG sin(1000t)
y = 4 cos 2
2 KJ
= 2 [cos t + 1] sin (1000t) = [sin(1000 + 1)t + sin(1000–1)t] + 2 sin 1000t = sin1001t + sin 999t + 2 sin 1000t.
4. Two simple harmonic motions of same frequency ω but having displacements in two perpendicular directions act simultaneously on a particle:
Find the resultant motion for various values of the phase difference δ = α 1 – α 2 .
Solution
We have
= sin ωt cos α + cos ωt sin α ,
b 2 ...(2.17) Multiplying Eqn. (2.16) by sin α 2 and Eqn. (2.17) by sin α 1 and subtracting the second from the first, we get
Similarly multiplying Eqn. (2.16) by cos α 2 and Eqn. (2.17) by cos α 1 and subtracting the second from the first, we obtain
Now squaring Eqns. (2.18) and (2.19) and adding, we obtain
x 2 y 2 2 xy
cos(α y – α (α – α
This represents the general equation of an ellipse. Thus, due to superposition of two simple A D harmonic vibrations at right angles to each other,
the displacement of the particle will be along a curve given by Eqn. (2.20).
Special Cases
(i) δ = α 1 – α 2 = 0, 2π, 4π,...
cos δ = 1, sin δ = 0 and
Fig. 2.3
SUPERPOSITION PRINCIPLE AND COUPLED OSCILLATIONS 63
− I = 0 or, y = b x. HG a b KJ
The particle vibrates simple harmonically along the straight line BD (Fig. 2.3).
(ii) δ = π, 3π, 5π,... We have
b y=– x
a This equation represents a straight line with
slope = – b/a. The particle vibrates along the straight line AC (Fig. 2.4)
x We have 2 y 2
2 a + b 2 =1
which is an ellipse with semimajor and semiminor axes a and b, coinciding with the x- and y-axes, respectively (Fig. 2.5). If a = b, we get the equation
of the circle x 2 +y 2 =a 2 with radius a.
We have
a b ab 2
This is an oblique ellipse (Fig. 2.6). a
x 2 y 2 2xy 1 Fig. 2.6
We have now
b ab 2
We get the oblique ellipse (Fig. 2.7). The direction of rotation (clockwise or
anticlockwise) of the particle may be obtained form the x- and y-motions of the particle when t is increased gradually. How the path of the particle with direction changes as δ is increased gradually is shown in Fig. 2.8.
Fig. 2.7
64 WAVES AND OSCILLATIONS
That the two cases δ = π/2 and δ = 3π/2, although giving the same path, are physically different may be seen by graphical constructions. When δ = π/2, we have
x = a cos (ωt + α 2 ) y = b sin (ωt + α 2 )
When t = 0 x = a cos α 2 and y = b sin α 2 . π
When ωt + α 2 = , x = 0 and y = b.
2 When ωt + α 2 = π, x = – a and y = 0. Thus in this case the particle moves in the anticlockwise direction.
Similarly, it can be shown that the particle moves in the clockwise direction
3 π when δ =
5. A particle is subjected to two SHMs represented by the following equations
x=a 1 sin ω t, y=a 2 sin (2ω t + δ )
in a plane acting at right angles to each other. Discuss the formation of Lissajous’ figures due to superposition of these two vibrations.
SUPERPOSITION PRINCIPLE AND COUPLED OSCILLATIONS 65 Solution
We have = sin 2ωt cos δ +cos 2ωt sin δ
1 − sin ω t cos δ + 1 −
a 1 HG a 1 2 KJ
2 HG a 1 KJ
Squaring this expression, we get
HG sin δ
− sin δ
a 2 KJ a 1 HG a 1 a 2 KJ
a 1 HG a 1 a 2 KJ HG a 2 KJ
This gives the general equation of the resultant motion for any phase difference and amplitudes.
Special Cases π
(i) When δ = , Eqn. (2.21) reduces to
−+ 1 2 HG =0
a 2 a 1 KJ
where represents two coincident parabolas:
The curve given by Eqn. (2.22) is shown in Fig. 2.9.
Fig. 2.9
66 WAVES AND OSCILLATIONS
(ii) When δ = 0, Eqn. (2.21) reduces to
a 2 a 1 HG a 1 KJ
This is an equation of 4th degree in x and it represents a curve having two loops (Fig. 2.10)
In Fig. 2.10,
y = 0 when x = 0, + a 1
and
y=+ a
2 when x = +
As the phase difference is changed gradually, the shape of the loop also changes gradually. Fig. 2.10 give the Lissajous’ figure for two simple harmonic vibrations in phase (δ = 0) with
a frequency ratio of 1:2 [frequency of x-vibration: frequency of y-vibration = 1 : 2].
Fig. 2.10
6. Two vibrations of frequencies in the ratio 1 : 3 and initial phase difference δ, given by
x=a 1 sin ω t, y=a 2 sin (3ω t + δ)
act simultaneously on a particle at right angles to each other. Find the equation of the figure traced by the particle.
Solution
We have = (3 sin ωt – 4 sin 3 ωt) cos δ + (4 cos 3 ωt – 3 cos ωt) sin δ
MM a
2 HG a 1 a 1 KJ Q PP
or
cos δ