Samacheer Class 11 Physics - Oscillations

(b) T’ = \(\frac{\mathrm{T}}{\sqrt{2}}\)
Hint:
T = 2π\(\sqrt{\frac{\mathrm{m}}{k}}\)
When the spring is cut into two equal halves, then the force constant of each part is 2k.
When the mass is suspended from one of the halves.
Time period T’ = 2π\(\sqrt{\frac{\mathrm{m}}{2k}}\)
= \(\sqrt{\frac{\mathrm{T}}{2}}\)

(b) \(\frac{d^{2} y}{d t^{2}}\) + γ \(\frac { dy }{ dt }\) + y = 0
Hint:
For a damped oscillator F ∝ v
Total restoring force F = – ky – bv
b – damping constant; y – displacement
If F = – ky – by
then m\(\frac{d^{2} y}{d t^{2}}\) = – ky – b\(\frac { dy }{ dt }\)
m\(\frac{d^{2} y}{d t^{2}}\) + ky + b\(\frac { dy }{ dt }\) = 0
÷ m we get
\(\frac{d^{2} y}{d t^{2}}\) + \(\frac { b }{ m }\)\(\frac { dy }{ dt }\) + \(\frac { k }{ m }\)y = 0
If k = m and \(\frac { b }{ m }\) = r
then the equation becomes
\(\frac{d^{2} y}{d t^{2}}\) + r\(\frac { dy }{ dt }\) + y = 0

Periodic motion: Any motion which repeats itself in a fixed time interval is known as periodic motion. Examples: Hands in a pendulum clock, the swing of a cradle.
Non-Periodic motion: Any motion which does not repeat itself after a regular interval of time is known as non-periodic motion. Example: Occurrence of Earthquake, the eruption of a volcano.

The number of oscillations produced by the particle per second is called frequency. It is denoted by f. SI unit for frequency is s -1 or hertz (Hz).
Mathematically, frequency is related to time period by f = \(\frac{1}{\mathrm{T}}\)

Initial phases of a particle is an epoch. At time t = 0 s (initial time), the phase φ = (φ 0 is called epoch (initial phase) where φ 0 is called the angle of epoch.

Let us consider two springs whose spring constant are k 1 and k 2 and which are connected to a mass m as shown in Figure.
Let F be the applied force towards right as shown in Figure. The net displacement of the mass point is
x = x 1 + x 2
From Hooke’s law, the net force,
F = – k s (x 1 + x 2 )
The effective spring constant can be calculated as
When n springs connected in series, the effective spring constant in series is
If all spring constants are identical
i.e., k 1 = k 2 = … = k n = k then
\(\frac{1}{k_{s}}\) = \(\frac { n }{ k }\) ⇒ k s = \(\frac { k }{ n }\)
This means that the effective spring constant reduces by the factor n. So, for springs in series connection, the effective spring constant is lesser than the individual spring constants.

Let us consider only two springs of spring constants k 1 and k 2 attached to a mass m as shown in Figure.
Net force for the displacement of mass m is F = – k p x … (1)
Where k p is called effective spring constant. Let the first spring be elongated by a displacement x due to force F 1 and the second spring be elongated by the same displacement x due to force F 2, then the net force
F = – k 1 x – k 2 x … (2)
Equating equations (2) and (1), we get
k p = k 1 + k 2 … (3)
Generalizing, for n springs connected in parallel,
k p = \(\sum_{i=1}^{n} k_{i}\) … (4)
If all spring constants are identical i.e., k 1 = k 2 = … = k n = k then
k p = nk … (5)
It is implied that the effective spring constant increases by a factor n. So, for the springs in parallel connection, the effective spring constant is greater than the individual spring constant.

Laws of simple pendulum: The time period of a simple pendulum.
Depends on the following laws:
(i) Law of length: For a given value of acceleration due to gravity, the time period of a simple pendulum is directly proportional to the square root of length of the pendulum.
T ∝ \(\sqrt{l}\) … (1)
(ii) Law of acceleration: For a fixed length, the time period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity.
T ∝ \(\frac{1}{\sqrt{g}}\) … (2)

The oscillations in which the amplitude decreases gradually with the passage of time are called damped Oscillations.
Example:
* The oscillations of a pendulum or pendulum oscillating inside an oil-filled container.
* Electromagnetic oscillations in a tank circuit.
* Oscillations in a dead beat and ballistic galvanometers.

If an oscillator oscillates in a resistive medium, then its amplitude goes on decreasing. The motion of the oscillator is called damped oscillation.
Example:
* The oscillations of a pendulum (including air friction) or pendulum oscillating inside an oil-filled container.
* Electromagnetic oscillations in a tank circuit.

The body executing vibration initially vibrates with its natural frequency and due to the presence of external periodic force, the body later vibrates with the frequency of the applied periodic force. Such vibrations are known as forced vibrations.
Example: Soundboards of stringed instruments.

To avoid damping in an oscillation, energy is supplied from an external source, the amplitude of the oscillation can be made constant. Such vibrations are known as maintained vibrations.
Example:
The vibration of a tuning fork getting energy from a battery or from an external power supply.

The frequency of external periodic force (or driving force) matches with the natural frequency of the vibrating body (driven). As a result, the oscillating body begins to vibrate such that its amplitude increases at each step and ultimately it has a large amplitude. Such a phenomenon is known as resonance and the corresponding vibrations are known as resonance vibrations. Example: The breaking of glass due to sound
III. Long Answers Questions:

Simple harmonic motion is a special type of oscillatory motion in which the acceleration or force on the particle is directly proportional to its displacement from a fixed point and is always directed towards that fixed point.
Example: Oscillation of a pendulum. SHM is a special type of periodic motion, where restoring force is proportional to the displacement and acts in the direction opposite to displacement.

(i) Let us consider a particle of mass m moving with uniform speed v along the circumference of a circle whose radius is r in an anti-clockwise direction (as shown in Figure).
(ii) It is assumed that the origin of the coordinate System coincides with the center 0 of the circle.
(iii) If co is the angular velocity of the particle and 0 the angular displacement of the particle at any instant of time t, then θ = ωt. By projecting the uniform circular motion on its diameter a simple harmonic motion is obtained.
(iv) This means that we can associate a map (or a relationship) between uniform circular (or revolution) motion to vibratory motion.
(v) Conversely, any vibratory motion or revolution can be mapped to uniform circular motion. The position of a particle moving is projected on to its vertical diameter or on to a line parallel to vertical diameter.
(vi) Similarly, we can do it for horizontal axis or a line parallel to the horizontal axis.
Example: Let us consider a spring-mass system (or oscillation of pendulum) as shown in Figure. When the spring moves up and down (or pendulum moves to and fro), the motion of the mass or bob is mapped to points on the circular motion.
Thus, if a particle undergoes uniform circular motion then the projection of the particle on the diameter of the circle (or on a line parallel to the diameter) traces straight line motion that is simple harmonic in nature. The circle is known as the reference circle of the simple harmonic motion.

If a body is allowed to rotate freely about a given axis then the oscillation is known as the angular oscillation. The point at which the resultant torque acting on the body is taken to be zero. It is called mean position. If the body is displaced from the mean position, then the resultant torque acts such that it is proportional to the angular displacement. This torque has a tendency to bring the body towards the mean position.
Time period: Let \(\overline{θ}\) be the angular displacement of the body and the resultant torque \(\vec { τ }\) acting on the body is,
\(\vec { τ }\) ∝ \(\vec {θ}\) … (1)
\(\vec { τ }\) = – k\(\vec {θ}\) … (2)
K is the restoring torsion constant, that is torque per unit angular displacement. If I is the moment of inertia of the body and \(\vec { τ }\) ∝ \(\vec {α}\) is the angular acceleration then
\(\vec { τ }\) = I \(\vec{α}\) = k \(\vec {θ}\)
But \(\vec{α}\) = \(\frac{d^{2} \vec{\theta}}{d t^{2}}\) and therefore
\(\frac{d^{2} \vec{\theta}}{d t^{2}}\) = – \(\frac { k }{ I }\)\(\vec {θ}\) … (3)
This differential equation resembles simple harmonic differential equation.
By comparing equation (3) with simple harmonic motion given we get,
a = \(\frac{d^{2} y}{d t^{2}}\) = – ω²y, we have
ω = \(\sqrt{\frac{k}{I}}\) rads -1 … (4)
The frequency of the angular harmonic motion is given equation ω = 2πf is
f = \(\frac { 1 }{ 2π }\) \(\sqrt{\frac{k}{I}}\)Hz … (5)
The time period is given equation is
T = \(\frac { 1 }{ f }\); T = 2π\(\sqrt{\frac{I}{k}}\) second
In angular simple harmonic motion, the displacement of the particle is measured in terms of angular displacement \(\vec {θ}\).

Construction:
A pendulum is a mechanical system which exhibits periodic motion. It has a bob with mass m suspended by a massless and inextensible string. The other end is fixed on a stand as shown in Figure (a). At equilibrium, the pendulum does not oscillate and is suspended vertically downward.
Such a position is known as mean position or equilibrium position. When a pendulum is displaced through a small displacement from its equilibrium position and released, the bob of the pendulum executes to and fro motion.
Calculation of time period: Let l be the length of the pendulum which is taken as the distance between the point of suspension and the centre of gravity of the bob. Two forces act on the bob of the pendulum at any displaced position, as shown in Figure (d),
* The gravitational force acting on the body (F = mg) acts vertically downwards.
* The tension in the string T acts along the string to the point of suspension.
Resolving the gravitational force into its components:
* Normal component: It is along the string but in opposition to the direction of tension, F as = mg cosθ.
* Tangential component: It is perpendicular to the string i.e., along the tangential direction of arc of the swing, F ps = mg sinθ.
Hence, the normal component of the force is, along the string,
T – W as = m\(\frac{v^{2}}{l}\)
Here v is speed of bob
T – mg cosθ = m\(\frac{v^{2}}{l}\) … (1)
From the Figure, it is observed that the tangential component W ps of the gravitational force always points towards the equilibrium position. This direction always points opposite to the direction of displacement of the bob from the mean position. Hence, in this case, the tangential force is the restoring force. Applying Newton’s second law along tangential direction, we have
Where, s is the position of bob that is measured along the arc. Expressing arc length in terms of angular displacement i.e.,
s = lθ … (3)
Then its acceleration,
\(\frac{d^{2} \theta}{d t^{2}}\) = \(\frac{d^{2} \theta}{d t^{2}}\) … (4)
Substituting equation (4) in equation (2), we get,
l\(\frac{d^{2} \theta}{d t^{2}}\) = – g sin θ
\(\frac{d^{2} \theta}{d t^{2}}\) = – \(\frac { g }{ l }\) sin θ … (5)
Because of the presence of sinθ in the above differential equation, it is a nonlinear differential equation. It is assumed that “the small oscillation approximation”, sin θ ≈ θ, the above differential equation becomes linear differential equation.
\(\frac{d^{2} \theta}{d t^{2}}\) = – \(\frac { g }{ l }\) θ
It is known as oscillatory differential equation. Hence, the angular frequency of this oscillator (natural frequency of this system) is

Let us consider a system containing a block of mass m fastened to a massless spring with stiffness constant or force constant or spring constant k placed on a smooth horizontal surface (frictionless surface) as shown in Figure. Let x 0 be the equilibrium position or mean position of mass m when it is left undisturbed.
When the mass is displaced through a small displacement x towards right from its equilibrium position and then released, it will oscillate back and forth about its mean position x 0. Let F be the restoring force (due to stretching of the spring) that is proportional to the amount of displacement of block: For one-dimensional motion, we get
F ∝ x
F = – kx
where negative sign implies that the restoring force will always act opposite to the direction of the displacement. This equation is called Hooke’s law. It is noticed that, the restoring force is linear with the displacement (i.e., the exponent of force and displacement are unity). This is not always true. If we apply a very large stretching force, then the amplitude of oscillations becomes very large.
m\(\frac{d^{2} x}{d t^{2}}\) = – kx
\(\frac{d^{2} x}{d t^{2}}\) = – \(\frac { k }{ m }\)x … (1)
Comparing the equation (1) with simple harmonic motion equation a = \(\frac{d^{2} y}{d t^{2}}\) = – ω²y, we get
ω² = \(\frac { k }{ m }\)
which means the angular frequency or natural frequency of the oscillator is
ω = \(\sqrt{\frac{k}{m}}\)rad s -1 … (2)
The frequency of the oscillation is
f = \(\frac { ω }{ 2π }\) = \(\frac { 1 }{ 2π }\)\(\frac { k }{ m }\) Hertz … (3)
and the time period of the oscillation is
T = \(\frac { 1 }{ f }\) = 2π[/latex]\(\frac { m }{ k }\) seconds … (4)

Consider a massless spring with stiffness constant or force constant k attached to a ceiling as shown in Figure. Let the length of the spring before loading mass m be L. If the block of mass m is attached to the other end of the spring, then the spring elongates by a length l. Let F 1 be the restoring force due to the stretching of spring. Due to mass m, the gravitational force acts vertically downward. A free-body diagram is drawn for this system as shown in Figure. When the system is under equilibrium,
F 1 + mg = 0 … (1)
But the spring elongates by small displacement l,
∴ F 1 ∝ l ⇒ F 1 = – kl … (2)
Substituting equation (2) in equation (1), we get
– kl + mg = 0
mg = kl
(or) \(\frac { m }{ k }\) = \(\frac { 1 }{ g }\) … (3)
Suppose a very small external force is applied on the mass such that the mass further displaces downward by a displacement y, then it will oscillate up and down. Now, the restoring force due to this stretching of spring (total extension of spring is y + l) is
F 2 (y + 1)
F 2 = – k(y + 1) = – ky – kl … (4)
Since, the mass moves up and down with acceleration, by drawing the free body diagram for this case, we get
-ky – kl + mg = m\(\frac{d^{2} y}{d t^{2}}\) … (5)
The net force acting on the mass due to this stretching is
F = F 2 + mg
F = – ky – kl + mg … (6)
The gravitational force opposes the restoring force. Substituting equation (3) in equation (6), we get
F = – ky – kl + kl = ky
Time period:
Applying Newton’s law, we get
m\(\frac{d^{2} y}{d t^{2}}\) = – ky
\(\frac{d^{2} y}{d t^{2}}\) = – \(\frac { k }{ m }\)y … (7)
The above equation is in the form of simple harmonic differential equation. Hence the time period is
T = 2π \(\sqrt{\frac{m}{k}}\)second … (8)
The time period can be rewritten using equation (3) as
T = 2π \(\sqrt{\frac{m}{k}}\) = 2π \(\sqrt{\frac{l}{g}}\)second.

Let us consider a U-shaped glass tube which consists of two open arms with uniform cross¬sectional area A. Let us pour a non-viscous uniform incompressible liquid of density p in the U-shaped tube to a height h as shown in the Figure. If the liquid and tube are not disturbed then the liquid surface will be in equilibrium position O.
It means the pressure as measured at any point on the liquid is the same and also at the surface on the arm (edge of the tube on either side), that balances the atmospheric pressure. Hence, the level of liquid in each arm will be the same. By blowing air one can provide sufficient force in one arm, and the liquid gets disturbed from equilibrium position O, It is meant, that pressure at blown arm is higher than the other arm.
A difference in pressure is created that will cause the liquid to oscillate for a very short duration of time about the mean or equilibrium position. Finally, it comes to rest.
Time period of the oscillation is
T= 2π \(\sqrt{\frac{l}{2g}}\)second

(i) Expression for Potential Energy:For the simple harmonic motion, the force and the displacement are related by Hooke’s law
\(\vec { F }\) = – k\(\vec { r }\)
F = – kx … (1)
the work done by the conservative force field is independent of path.
Calculation potential energy:
F = \(\frac { dU }{ dx }\) … (2)
Comparing (1) and (2), we get
\(\frac { dU }{ dx }\) = – kx
dU = – kx dx
This work done by the force F during a small displacement dx stores as potential energy
From equation,
ω = \(\sqrt{\frac{k}{m}}\) rad s -1
By substituting the value of force constant k = mω² in equation (3), we get
U(x) = \(\frac { 1 }{ 2 }\)mω²x² … (4)
where ω is the natural frequency of the oscillating system. For the particle executing simple harmonic motion from the equation,
y = A sin ωt
we get,
x = A sin ωt
U(t) = \(\frac { 1 }{ 2 }\)mω²x² … (5)
This variation of U is shown below.
(ii) Expression for Kinetic Energy:
Kinetic energy
KE = \(\frac { 1 }{ 2 }\)mv x ² = \(\frac { 1 }{ 2 }\)m(\(\frac { dx }{ dt }\))² … (6)
Since the particle executes simple harmonic motion, from equation y = A sin ωt
x = A sin ωt
∴ velocity is
(iii) Expression for Total Energy: Total energy is the sum of kinetic energy and potential energy.
E = KE + U … (11)
E = \(\frac { 1 }{ 2 }\)mω²(A² – x²) + \(\frac { 1 }{ 2 }\)mω²x² … (12)
Hence, cancelling x² term,
E = \(\frac { 1 }{ 2 }\)mω²x² = r constant … (13)
Alternatively, from equation (5) and equation (10), we get the total energy as
E = \(\frac { 1 }{ 2 }\)mω²x² sin²ωt + \(\frac { 1 }{ 2 }\)mω²A²cos²ωt
= \(\frac { 1 }{ 2 }\)mω²A² (sin²ωt + cos²ωt)
From trigonometry identity,
(sin² ωt + cos² ωt) = 1
E = \(\frac { 1 }{ 2 }\)mω²A² = constant
which gives the law of conservation of total energy.

(i) Free oscillations: When the oscillator oscillates with a frequency that is equal to the natural frequency of the oscillator. Such an oscillation or vibration is known as free oscillation or free vibration.
Example:
* Oscillation of a simple pendulum,
* Vibration in a stretched string.
(ii) Damped oscillation: If an oscillator oscillates in a resistive medium, then its amplitude goes on decreasing. The energy of the oscillator is used to do work against the resistive medium. The motion of the oscillator is said to be a damped oscillation.
Example:
* The oscillations of a pendulum (including air friction) or pendulum oscillating inside an oil filled container.
* Electromagnetic oscillations in a tank circuit.
(iii) Forced oscillations: In this type of vibration, the body executing vibration initially vibrates with its natural frequency. Because of the presence of external periodic force, the body later vibrates with the frequency of the applied periodic force. Such vibrations are known as forced vibrations.
Example:
Soundboards of stringed instruments.
(iv) Maintained oscillations: The amplitude of the oscillation can be made constant. By supplying energy from an external source. Such oscillations are known as maintained oscillations.
Example:
The vibration of a tuning fork getting energy from a battery or from external power supply.
IV. Numerical Problems:

In the problem consider oscillation of particle into 2 cases via x < 0 (consider it as SHM where the time period is considered as t 1 ) and another one as x > 0 (consider it as a motion under gravity, where time period is t 2 )
Note:
We want to find the total time period, which will be T = t 1 + t 2.
According to the conservation of energy
In case: 2 (Motion under gravity)
t 2 = \(\frac { 2v }{ g }\) substituting equation (2) here we get,
t 2 = \(\frac { 2 }{ g }\)\(\sqrt{\frac{2 E}{m}}\)
⇒ 2\(\sqrt{\frac{2 \mathrm{E}}{m g^{2}}}\) … (4)
Adding equation (3) & (4)
Time period of oscillation,
T = t 1 + t 2
= π\(\sqrt{\frac{m}{k}}\) + 2\(\sqrt{\frac{2 \mathrm{E}}{m g^{2}}}\)

The effective value of acceleration due to gravity (g) will be equal to the component of g normal to the inclined plane which is
g’ = g cos α
T = 2π\(\sqrt{\frac{l}{g^{\prime}}}\) = 2π\(\sqrt{\frac{l}{g \cos \theta}}\)
Length of the pendulum l = 0.9m
Angle of inclination θ = 45°

(a) Suppose a particle of mass m executes SHM of time period T. The displacement of the particle at any instant t is given by
y = A sin ωr … (1)
Velocity v = \(\frac { dy }{ dt }\) = \(\frac { d }{ dt }\)(sin ωt) = Aω cos ωt = ωA cosωt
Average potential energy over a period of oscillation is,
(b) Total energy
T.E = \(\frac { 1 }{ 2 }\) ω²y² + \(\frac { 1 }{ 2 }\)mω²(A² – y²)
But y = A sinωt
T.E = \(\frac { 1 }{ 2 }\)mω²A² ωt + \(\frac { 1 }{ 2 }\)mω²A² cos²ωt
= \(\frac { 1 }{ 2 }\)mω²A²(sin²ωt + cos²ωt)
From trignometry identity
sin² ωt + cos² ωt = 1

Case (a):
When mass is slightly displaced vertically down: Now pulley is fixed rigidly here. When the mass is displaced by y and the spring will also be stretched by y.
Hence F = ky
Time period T = \(\sqrt{\frac{m}{k}}\)
case (b):
When the system is released: WTien mass is displaced by y, pulley is also displaced by 4y,
∴ F = 4 Icy
∴ T = 2π\(\sqrt{\frac{m}{4k}}\)