**Introduction**

In this chapter we will learn about oscillatory motion or oscillations. Any motion which repeats itself at regular intervals of time is known as periodic motion. If a body moves back and forth repeatedly about its mean position then it is said to be in oscillatory motion.

For example: The to and fro movement of pendulum, jumping on a trampoline, a child swinging on a swing.

Oscillations can be defined as Periodic to and fro motion which repeat itself at regular intervals of time.

__Periodic and Oscillatory motions__

Oscillations are defined as to and fro motion which repeat itself after regular intervals of time.In oscillations, the frequency of vibrations iscomparatively less.

For example: The to and fro motion of a pendulum clock

__Frequency__

It is defined as number of cycles per second.

- It is denoted by ν.
- I.unit is sec
^{-1} - Special Unit is Hertz(Hz)

** Problem**: – On an average human heart is found to be beat 75 times in a minute. Calculate frequency and time period?

** Answer**: –

The beat frequency of heart = 75/ (1 min)

= 75/ (60 s)

= 1.25 s^{–1}

= 1.25 Hz

The time period T = 1/ (1.25 s^{–1}) = 0.8 s

In the above image we can see that there is a block whose one end is attached to a spring and another is attached to a rigid wall.x is the displacement from the wall.

In the above figure a blockis attached to a spring, the other end of which is fixed to a rigid wall. The block moves on a frictionless surface. The motion of the block can be described in terms of its distance or displacement x from the wall.

**f (t) = A cos ωt**

As cosine function repeats after 2π so it can be written as

cos (θ) = cos (ωt + 2π) Equation (1)

cos (ωt) = cos (ωt + 2π) (it keep on repeating after 2π)

Let Time Period = T

f (T) = f(t+T) where displacement keeps on repeating after (t+T)

Acos (ωt) = cosω(t+T) = Acos (ωt+ wT)

Acosωt = A cos (ωt+ωT) Equation (2)

From Equation (1) and Equation (2)

ωT= 2π

Or **T=2π /ω**

__Displacement as a combination of sine and cosine functions__

f (t) = A cos ωt

f (t) = A sin ωt

f (t) = A sin ωt + A cos ωt

LetA = D cosΦ Equation (3)

B=DsinΦEquation (4)

f (t) =DcosΦ sinωT + DsinΦ cos ωt

D (cosΦ sinωT + sinΦ cos ωt)

(Using sinAcosB + sinBcosA = sin (A+B))

Therefore we can write

f (T)= **D sin (ωT+****Φ)**

From the above expression we can say displacement can be written as sine and cosine functions.

__D in terms of A and B:-__

A^{2 }B^{2 }= D^{2}sin^{2} Φ + D^{2}cos^{2} Φ

A^{2 }B^{2 }= D^{2}

Or D= AB

Φ In terms of A and B

Dividing Equation (4) by (3)

B/A= DsinΦ/Dcos Φ

tan Φ = B/A

Or Φ= tan^{-1} B/A

** Problem:-**Which of the followingfunctions of time represent (a) periodic and (b) non-periodic motion? Give the period foreach case of periodic motion [ω is anypositive constant].

(i) sin ωt + cos ωt

(ii) sin ωt + cos 2 ωt + sin 4 ωt

(iii) e–ωt

(iv) log (ωt)

__Answer:-__

- sin ωt + cos ωt is a periodic function, it can also be written as

2 sin (ωt + π/4).

Now 2 sin (ωt + π/4)= 2 sin (ωt + π/4+2π)

= 2 sin [ω (t + 2π/ω) + π/4]

The periodic time of the function is 2π/ω.

(ii) This is an example of a periodic motion. Itcan be noted that each term represents aperiodic function with a different angularfrequency. Since period is the least intervalof time after which a function repeats itsvalue, sin ωt has a period T_{0}= 2π/ω; cos 2 ωt

has a period π/ω =T_{0}/2; and sin 4 ωt has aperiod 2π/4ω = T_{0}/4. The period of the firstterm is a multiple of the periods of the last

two terms. Therefore, the smallestintervalof time after which the sum of the threeterms repeats is T_{0}, and thus the sum is aperiodic function with aperiod 2π/ω.

(iii) The function e^{–ωt} is not periodic, itdecreases monotonically with increasingtime and tends to zero as t →∞ and thus,never repeats its value.

(iv) The function log (ωt) increases monotonicallywith time t. It, therefore, neverrepeats its value and is a non-periodicfunction. It may be noted that as t →∞,log (ωt) diverges to ∞. It, therefore, cannot represent any kind of displacement.

__Phase__

It is that quantity that determines the state of motion of the particle.

- Its value is (ωt + Φ)
- It is dependent on time.

Value of phase at time t=0, is termed as **Phase Constant**. When the motion of the particle starts it goes to one of the extreme position at that time phase is considered as 0.

Let x (t) = A cos (ωt) where we are taking (Φ = 0)

**Mean Position**(t= 0)- x (0) = A cos (0) = A (cos0=1)
- t=T/4, t= T/2, t=3T/4, t=T and t=5T/4

** Answer**:

In SHM, acceleration a is related to displacement by the relation of the form

a = -kx, which is for relation (c).

** Problem**: – The motion of a particle executing simple harmonic motion is described by the displacement function,

x (t) = A cos (ωt + φ).

If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s^{–1}. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

** Answer**:-

Initially, at t = 0;

Displacement, x = 1 cm

Initial velocity, v = ω cm/ sec.

Angular frequency, ω = π rad/s^{–1}

It is given that,

x (t) = A cos (ωt + Φ)

1 = A cos (ω × 0 + Φ) = A cos Φ

A cosΦ = 1 … (i)

Velocity, v= dx/dt

ω = -A ω sin (ωt + Φ)

1 = -A sin (ω × 0 + Φ) = -A sin Φ

A sin Φ = -1 … (ii)

Squaring and adding equations (i) and (ii), we get:

A^{2} (sin^{2} Φ + cos^{2} Φ) = 1 + 1

A^{2} = 2

∴ A = √2 cm

Dividing equation (ii) by equation (i), we get:

tan Φ = -1

∴Φ = 3π/4, 7π/4…

SHM is given as:

x = B sin (ωt + α)

Putting the given values in this equation, we get:

1 = B sin [ω × 0 + α] = 1 + 1

B sin α = 1 … (iii)

Velocity, v = ω B cos (ωt + α)

Substituting the given values, we get:

π = π B sin α

B sin α = 1 … (iv)

Squaring and adding equations (iii) and (iv), we get:

B2 [sin2 α + cos2 α] = 1 + 1

B2 = 2

∴ B = √2 cm

Dividing equation (iii) by equation (iv), we get:

B sin α / B cos α = 1/1

tan α = 1 = tan π/4

∴α = π/4, 5π/4…

__Angular Frequency (____ω)__

Angular frequency refers to the angular displacement per unit time. It can also be defined as the rate of change of the phase of a sinusoidal waveform (e.g., in oscillations and waves). Angular frequency is larger than frequency ν (in cycles per second, also called Hz), by a factor of 2π.

Consider the oscillatory motion which is varying with time t and displacement x of the particle from the origin:

x (t) = cos (ωt + Φ)

Let Φ = 0

x (t) = cos (ωt)

After t=T i.e. x (t) = x (t+T)

A cos ωt = A cos ω (t + T)

Now the cosine function is periodic with period 2π, i.e., it first repeats itself after 2π. Therefore,

ω (t + T) = ωt + 2π

i.e. **ω = 2π/ T**

Where ω = angular frequency of SHM.

- I. unit is radians per second.
- It is 2π times the frequency of oscillation.
- Two simple harmonic motions may have the same A and φ, but different ω.

** Problem**: – Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant):

(a) sin ωt – cos ωt

(b) sin3ωt

(c) 3 cos (π/4 – 2ωt)

(d) cos ωt + cos 3ωt + cos 5ωt

(e) exp (–ω^{2}t^{2})

** Answer**: –

(a) SHM

The given function is:

sinωt – cos ωt

This function represents SHM as it can be written in the form: a sin (ωt + Φ)

Its period is: 2π/ω

(b) Periodic but not SHM

The given function is:

sin 3ωt = 1/4 [3 sin ωt – sin3ωt]

The terms sin ωt and sin ωt individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.

Its period is: 2π/ω

(c) SHM

The given function is:

This function represents simple harmonic motion because it can be written in the form: a cos (ωt + Φ) its period is: 2π/2ω = π/ω

(d) Periodic, but not SHM

The given function is cosωt + cos3ωt + cos5ωt. Each individual cosine function represents SHM. However, the superposition of three simple harmonic motions is periodic, but not simple harmonic.

(e) Non-periodic motion

The given function exp (-ω^{2}t^{2}) is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.

(f) The given function 1 + ωt + ω^{2}t^{2} is non-periodic.

__Problem__**: **Which of the followingfunctions of time represent (a) simple harmonic motion and (b) periodic but notsimple harmonic? Give the period for each case?

(1) sin ωt – cos ωt

(2) sin^{2} ωt

__Answer__**:**

(a) sin ωt – cos ωt

= sin ωt – sin (π/2 – ωt)

= 2 cos (π/4) sin (ωt – π/4)

= √2 sin (ωt – π/4)

This function represents a simple harmonicmotion having a period T = 2π/ω and aphase angle (–π/4) or (7π/4).

(b) sin^{2} ωt

= ½ – ½ cos 2 ωt

The function is periodic having a period T = π/ω. It also represents a harmonic motion with the point of equilibrium occurring at ½ instead of zero.