Oscillations:: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514 1
Oscillations:: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514 1
Oscillations:: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514 1
: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514
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DISPLACEMENT IN S.H.M. Sp.Case.(a) At mean position y = 0 & ωt =0
The displacement of a particle executing vmax=aω
S.H.M at an instant is defined as the (b) At extreme position y = ±a & ωt = π/2
distance of particle from the mean position =
at that instant. ACCELERATION IN S.H.M
A= = (aωcosωt)
NOTE:
∴ T = 2π
: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514
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(b) From equation ω = k/m
( )
∴ T = 2π
( )
: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514
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Motion of a simple pendulum becomes equal to its initial value when
sphere is empty.
(c) Effect of g:
T∝ i.e., g increase T decrease.
As we go high above the earth’s surface or
we go deep inside the mines the value of g
decreases, hence time period of pendulum
Restoring force, F = -mg sinθ = -mg θ (if θ (T) increases.
is small and measured in radians)
F = - mg. = − . ……….(1)
∴ F ∝ -y , Hence motion is S.H.M
But F = −mω y …………….(2)
Comparing equation (1) & (2) we get,
: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514
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Vρg = Vρg − Vσg ⇒ = l
T= , T will decrease.
, 2 1/2
⇒ = = >1 g2 + qE
m
(2) Oscillation under the influence of
(3) Pendulum in a lift: If pendulum is
electric field: If a bob of mass m carries
suspended from the ceiling of the lift.
a positive charge q and pendulum is
(i) If the lift is at rest or moving upward
placed in uniform electric field of strength
with constant velocity.
E.
(i) If electric field is directed vertically
upwards.
Effective acceleration g = g − T = 2π , Time period unchanged.
T = 2π , Time period
decreases.
(iii) If the lift is moving downward with
constant acceleration a.
T = 2π , Time period
g =g+ ,
T = 2π , T will decrease.
g = g +a ⇒ T = 2π ( ) /
And θ = tan
If simple pendulum suspended in a car
that is moving with constant speed v
around a circle of radius r.
T = 2π
g = g + ,
: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514
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SOME OTHER TYPES OF PENDULUM Comparing with equation = −ω θ, we get
(1) Infinite length pendulum.
If length of the pendulum is comparable = ⇒ T = 2π Also = I + ml
to the radius of earth then
= mk + ml ( where k = radius of gyration )
∴ =2 = 2π = 2π
: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514
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(2) S.H.M of a floating cubical block : B = bulk modulus of elasticity of air,
Let h be the immersed depth of cube m = mass of the ball.
initially. Ah = displaced volume Consider the ball be displaced from B1 to
B2 where , B1B2 = y. As a result of this,
air inside the chamber is compressed. If
∆V is the decreased in volume,
∆V = Volume of air column of length B1B2
= Ay
T = 2π
Let , A = area of cross-section of neck,
V = Volume of the air chamber, Where R= radius of the bowl
Ρ = density of air in chamber, r = radius of the ball
: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514
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(5) S.H.M of piston in a cylinder.
: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514
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LISSAJOUS FIGURE IN OTHER
CONDITIONS =
+ − =0⇒ − =0⇒y= x
The path is a straight line segment (see
figure) tan θ =
This motion in SHM with amplitude
A +A
Case (ii) α = π, In this case, we have
+ + =0 ⇒ =− x
The path is a straight line segment
inclined at and θ where the amplitude is
A +A
Case (iii) = /2 In this case, we have For the frequency ratio 1 : 2 = 2 : 1
the two perpendicular SHM’s are
= a sin( + ∅) = sin
Different Lissajous figure are as follows
+ =1 , This is an ellipse.
Case (iv) α = π/2, = In this case we
have,
SUPERPOSITION OF SHM IN SAME
DIRECTION
Let two SHMs of equal frequency be acting
on a body along the same line
x + y = A , This is a cicle.
: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514
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x = A sinωt & x = A sin(ωt + α) If F1 be the restoring force set up in the
= A sinωt + A sin(ωt + α) spring, then F1 = -kl
using the rotation vector A ⃗and A ⃗, we have { As the system is in equilibrium
F1 + mg = 0 ∴ = }
If F2 is the restoring force set up in the
spring, when it is pulled through a small
distance y, then F2 = -k ( l + y )
So, effective restoring force
F = F1 + F2 = - k ( l + y ) - kl = -ky …….(1)
But F = -mω x …………….(2)
Equating (1) and (2) we get =
∴ x = A ⃗. ȷ̂ and x = A ⃗. ȷ̂ , = A ⃗. ȷ̂ + A ⃗. ȷ̂ Or T = 2π
= A ⃗ + A ⃗ ȷ̂ = Asin(ωt + δ)
A = A + A + 2A A cosα Puttin the value of k we get, T = 2π
Where tan δ =
The time period of a loaded massless
SPRING MASS SYSTEM spring is equal to that of a simple
(i) Horizontal position. pendulum whose length is equal to the
When mass is displaced through a small extension in the spring.
distance x, a restoring force F = -kx …(1) COMBINATION OF SPRING.
When the mass is released it start (i) Series combination. If two springs of
oscillating back and forth about the spring constants K1 and K2 are joined in
equilibrium position under the influence of series as shown then
this restoring force.
: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514
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acts on different spring but extension in (iv) The force pruducing a resistance to the
springs are same. oscillation is called damping force.
(b) K = K + K If the velocity of oscillator is v then
Damping force
MASSIVE SPRING : F = −bv, b = damping constant.
If the spring has a mass M and mass m is (v) Resultant force on damping oscillator is
suspended from it, effective mass is given given by F = + =− −
by m =m+ . Hence T = 2π ⇒ + + =
/
x = and
( )
: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514
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( ) The length of the pendulum B and D are
= equal. The pendulum A is shorter, whereas
/
the pendulum C is longer than the
where ω = = Natural frequency. pendulums B and D.
RESONANCE. When the frequency of Displaced the bob of the pendulum D to
external force is equal to the natural one side and release it so as to set it into
frequency of the oscillator. Then this state vibrations. These vibrations are
is known as the state of resonance. And communicated to all other pendulums
this frequency is known as resonant through rubber string and as a result, all
frequency. of them start vibrating. The vibrations of A
AMPLITUDE RESONANCE . The and C are irregular to start with but after
amplitude of forced oscillator depend upon some time these settle to vibrate with the
the frequency ω of external force. frequency of D. The amplitude of
When nω = ω , the amplitude is vibrations of B goes on increasing till it
maximum but not infinite because of becomes equal to that of D. Clearly, the
presence of damping force. The vibrations of D are free vibrations. Since
corresponding frequency is called resonant the pendulums b and D have the same
frequency (ω ) length, their natural time periods and
frequencies are the same. The vibrations of
B are, therefore, resonant vibrations. The
pendulum A and C, Which have diffreent
lengths, are made to vibrate with the
frequency of D, which is different from
their natural frequencies. Hence, the
vibrations of A and C are forced vibrations.
MAINTAINED OSCILLATION.
The oscillation in whics the loss of energy
of oscillator is compensated by the
supplying energy from an external source
are known as maintained oscillation.
Example :
(i) the balance wheel of a watch where the
main spring supplies the external energy.
(ii) Electrically maintained tuning fork
(iii) An electronic oscillator.
ILLUSTRATION OF FREE, FORCED AND
RESONMANT OSCILLATIONS.
Suspended four simple pendulums A, B, C
and D with light bobs from a rubber string
stretched between two fixed point P and Q
as shown in figure.
: F-13/5, SREET NO.1, NAFEES ROAD, JOGABAI EXT, NEW DELHI-110025.Ph-011 26988514
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