CLASS 11 OSCILLATION
OSCILLATION
✤ PERIODIC MOTION -> It is a motion in which a body repeat its initial position after fixed interval of time.
Ex: Motion of the earth around Sun.
✤ OSCILLATORY MOTION -> It is to-fro or back & forth motion of a Body about a fixed point.
Ex: Motion of pendulum about a fixed point
NOTE:- Every bounded & periodic motion is called Oscillatory.
→ Every Oscillatory is periodic but every periodic is not oscillatory.
✤ PERIODIC FUNCTION -> Those functions which give same values for different values of x out.
let f(x) is any function
If f(T+x) = f(x)
then f(x) is called periodic function & T is called Period.
For eg:- (i) f(x)= Sinx (ii) f(x) = Cosx
f(2π+x) = Sin(2π+x) Period = 2π
= Sinx
= f(x)
: Period = 2π
✤ HARMONIC MOTION -> Those oscillatory motion in which displacement of oscillating particle (body) is represented by single function of sine or cosine.
ω = θ / T ω = 2πν ω= Angular Frequency
θ = ωT ω = 2π/T
Ex: Check Whether y = Sinwt + Coswt is harmonic.
Sol: y = Sinwt + Coswt
y = √2 ( 1/√2 Sinwt + 1/√2 Coswt)
y = √2 (Cos(π/4) Sinwt + Sin(π/4) Coswt)
y = √2 (Sinwt x Cos(π/4) + Coswt. Sin(π/4))
y = √2 Sin (wt + π/4)
:. Yes, it is harmonic motion.
Ex: y = Coswt + Cos 2wt + Cos 6wt
y = ewt
y = log (1+wt)
No, it is Non Harmonic Motion
✤ Simple Harmonic Motion (SHM)
It is a special type of harmonic motion in which restoring force always acts towards the mean position.
In SHM
-F ∝ x
F = -kx
K = -F/x
Unit: Nm⁻¹
✤ Velocity in SHM
Let a particle of mass 'm' excite SHM with radius a. At any instant 't' P(x,y)
Sinθ = y/a
y=sin θ & Cos θ = x/a
y = asinwt x = a Cos θ
x= a Cost
Vertical displacement
Vertical Comp. of Velocity :-
Vy = dy/dt
= d(asinwt) / dt
= (aCoswt) . w
Vy = awCoswt
= aw √1-sin²wt
= aw √1-y²/a²
= aw √a²-y² / a
Vy = w√a²-y²
Horizontal Comp. of Velocity :-
At any instant
x = aCoswt
dx/dt = - aSinwt.w
Vx = - awSinwt
Vx = - aw√1-cos²wt
= -aw √1-x²/a²
Vx = -w√a²-x²
✤ TIME PERIOD IN SHM :-
We know
Acceleration = dVy/dt
= d/dt (a cosωt)
= aω . d(-cosωt)/dt
= aω (-sinωt).ω
= -ω² a sinωt
A = -ω²y
∴ F = MA = -mω²y
we have, f = -Ky
-Ky = -mω²y
k=mω²
=> ω² = k/m
=> ω = √k/m
=> 2π/T = √k/m Here m = Inertial factor
T = 2π √(m/k) K = Spring Constant
Energy in SHM => it is the sum of K.E & P.E
Now Energy = KE + PE
Here KE = 1/2 mv²
= 1/2 m (ω² a²- y²)
KE = 1/2 mω²(a²-y²)
Now, PE equal to the amount of Work done against restoring force
W = - f X displacement
dw= + ky X dy
dw = (mw² . y²) / 2
∴ PE= ½ mw2y²
Total energy = ½ mw (a² - y²) + ½ mw2y²
= ½ mw (a² - y² + y²)
TE = 1/2 mw2a²
✤ OTHER FORMULAS :-
KE = ½ mw²A² sin² (wt + Φ₀)
KE = ½ mw² (A² - x²) = ½ k(A² - x²)
PE= ½ kx² = ½ mw² A² cos² (wt + Φ₀)
At lowest or mean position ke is maximum i.e. ½ mw²A²
At lowest or mean position PE is zero,
✤ Phase: The phase of a vibrating at any instant gives the state of the particle as regards its position & the direction of motion at that instant.
𝜙x = wt + 𝜙₀
Initial Phase = Epoch (𝜙₀)
Maximum Velocity – Maximum acceleration, aₘₐₓ = ω²A
V = ωA a = ω²A
* At turning points, acceleration is maximum
[aₘₐₓ = ω²A = ug]
✤ SIMPLE PENDULUM
It consists a string (inextensible) & a metallic bulb is connected at its free end
Let a lobe of mass m is displaced at small angle '𝛳' from mean position
At extreme position
mg cos𝛳 = T(tension) ----- (1)
mg sin𝛳 is used as restoring force.
We know restoring torque
T = mg sin𝛳 . L -------(2)
for small displacement sin𝛳 ≈ 𝛳
T = mg𝛳 . L -----(3)
we know angular torque
T = -k𝛳, [k torque per unit elment]
from (1) & (2)
+k𝛳 = +mg𝛳L
k = mgL
We know, time period in SHM, T = 8π √(inertial factor / K)
T= 2π √ ml² / mgl
T= 2π √(l/g)
✤ FACTOR AFFECTING TIME PERIOD :-
(i) effect of length:
we know, T= 2π √(l/g)
If g = Constant
then T ∝ √l
see T1 / T2 = √L1 / √L2
(ii) Effect of temp :-
we know Δl ∝ Δt
& we know time period T ∝ √l.
It shows that with rise in temp: lengthy pendulum increases i.e time
period increase i.e Pendulum Clock get slower.
(iii) Effect of Gravity :-
We Know, T = 2π √ l / g
If L = Constant
T ∝ 1 / √g
It shows that on the mountains or inside the mines g decreases, So T increases i.e clock of pendulum get slower.
(iv) Effect of motion:-
(a) when the elevator is going upward with acceleration a then,
T α L/√(g+a)
(b) when the elevator is going down with acceleration 'a' then
T α 1/ √(g-a)
(c) In freefall a=g,
T=∞
(d) Motion in an inclined plane
g'=gcosθ
T α 1 / √gcosθ
✤ SECOND PENDULUM -> It is a simple pendulum which have time period 2 seconds.
we know, T= 2π √(l/g)
1/π= √(l/g)
1/π*π = l/g
L=g/pi = 9.8 / (3.14)2
L = 0.998 m
L = 1m (app.)
✤ SPRING
Let a spring of length l. is hanged from one of its end & Spring Constant k.
If a body of mass 'm' is hanged from its free end the restoring force
F₁=-kx -①
Now, spring is pulled upto 'y' distance & let it free to oscillate, then
F₂=-k(x+y) -②
Net restoring force f=f₂-f₁
=-kx-(kx)-ky
f=-ky
It shows that it make SHM
So time period in SHM
T=2π √(m/k)
✤ Grouping of Springs:-
→ Parallel Method: Let two springs of spring constant k₁ & k₂ are hanged from vertical. A body of mass 'm' is also hanged at its free end & free to oscillate then
f₁=-k₁y & f₂=-k₂y
Total restoring:
f=f₁+f₂
f=-(k₁+k₂)y
let k₁+k₂=K
F = -ky
It shows that motion is in SHM.
We know, T = 2π√(m/K_1K_2)
→ SERIES METHOD
Let two springs of spring Constant K_1 & K_2 are connected in series & due to weight mg. y_1 & y_2 are elongations!
So, F= -k1y1 ------ ① ⇒ Y1 = -F/K1
F= -k2y2 ------ ② ⇒ Y2 = -F/K2
Y1+Y2 = f(1/K1 + 1/K2)
y= -f(K1 + K2)/K1K2
f = (-(K1K2)/(K1 + K2)) y
.: f = -Ky
It shows SHM
∴ T = 2π √(m/(k1k2/(k1+k2)))
∴ T = 2π √(k1k2/(k1+k2) m)
✤ Question Tip :-
In a horizontal turntable,
Restoring Force = Centripetal force
F = Ky = -mrω² = mr(2πν)²
K = 4π²ν²mr / y
The Time period of a body suspended by a spring loc. T, if the spring is cut into two equal parts, the new time period,
T' = 2π √m/√2k = T/√2
• As V α 1/√L , so the no. of seconds gained per second day on decreasing the length by 2%.
δD = 1 δl/2l x 86400
• Oscillation of a liquid Column containing mercury :-
K = 2Ap g T = 2π√l/2g
• Oscillation of a body dropped in a tunnel along diameter of earth :-
K = mg/R T = 2π √m/mgR = 2π√R/g
• Oscillation of a floating cylinder :-
K = Aρ1g T = 2π√ρh/ρ1g
ρ₁ = density of water
ρ = density of cylinder
✤ FREE OSCILLATIONS: if a body, capable of oscillations, is slightly displaced from its mean position, then left to itself will start oscillations with a frequency of its own.
eg: The oscillations of the bob of a pendulum when displaced permits mean position & released.
✤ DAMPED OSCILLATIONS: the oscillations in which the amplitude decreases gradually with the passage of time are called damped oscillation.
eg: The oscillations of swing in air
✤ MAINTAINED OSCILLATIONS: If to an oscillating system, energy is continuously supplied from outside at same ratio at which energy is lost lay it & then its amplitude can be maintained.
eg: An electrically maintained tuning fork.
✤ SHORT QUESTIONS:
→ The rotation of earth about it axis is periodic not SHM because the basic concept of To & fro is missing.
→ Restoring force for SHM Oscillations Simple pendulum : Gravity Elasticity: Spring
Column of Hg in U-tube: Weight of difference Column.
→ Direction of Displacement & Velocity are same when it moves from mean position to extreme position, Direction of acc. & velocity are same
when particle moves from extreme position to mean position. Tension of the string is max at mean position
Tension of the string is min. at either extreme position.
→ Bob of simple pendulum is replaced by wooden bole then iron bob there will be no effect because time period does not depend upon the nature of Bob.
→ Time period also does not depend upon density.
→ When the frequency of the sound waves from the engine of an aero planes matches with the natural frequency of window, resonance take place which causes rattling of window.
Question Tip:-
Δf = 86400 x (Change / 200)
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