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Oscillations

Objectives: Oscillations

25 — Oscillations (Textbook-style Page)

Oscillations

INTRODUCTION

So far we have studied translational and rotational motion and how we have used them in examining several areas of physics. Another important kind of motion that we now introduce is oscillatory or vibratory motion. When an object moves back and forth repeatedly over the same path, it is said to be oscillating or vibrating.

An important type of oscillatory motion is simple harmonic motion (S.H.M.), for which displacement–time graphs form a sine curve across time. The motion of a mass bouncing up and down on a spring, the periodic tilting of a simple pendulum, or the vibration of a tuning fork are familiar examples. In studying simple harmonic motion we rely on Newton’s second law, kinematics and energy ideas to analyse this simple type of motion.

25.1 Periodic and Oscillatory Motion

Before discussing simple harmonic motion, it is desirable to define periodic motion and oscillatory motion.

1. Periodic motion

Definition. The motion which repeats itself after a regular interval of time is called periodic motion. The regular time interval is called the period of the periodic motion.

Examples. A few examples of periodic motion are given below:

The revolution of the Moon around the Earth is a periodic motion. Its period of revolution is about 27.3 days.
The revolution of the Earth around the Sun is a periodic motion. Its period is one year.
The motion of the hands of a clock is a periodic motion.

2. Oscillatory motion

Definition. Oscillatory motion is a periodic motion in which a body moves to and fro repeatedly on the same path about a fixed point called the mean or equilibrium position.

An object that oscillates is displaced from the equilibrium position; a restoring force acts on it that tries to pull it back towards equilibrium. If the restoring force is directly proportional to displacement and directed towards the equilibrium position, the motion is called simple harmonic motion.

Examples. A few examples of oscillatory motion are given below:

The to-and-fro motion of a simple pendulum is oscillatory.
The prongs of a tuning fork execute oscillatory motion.
A mass attached to a spring and pulled aside performs oscillations about its equilibrium position.
When a loaded spring is pulled and then released, the load attached to the spring executes oscillatory motion.

25.2 — Simple Harmonic Motion (S.H.M.)

25.2. Simple Harmonic Motion (S.H.M.)

A simple harmonic motion (S.H.M.) is an important type of oscillatory motion and may be defined as under:

A particle is said to execute S.H.M. if it moves to and fro about the mean (equilibrium) position in a straight line and the graph between displacement of the particle from equilibrium position and time is a sine curve (or cosine curve).

In simple harmonic motion, the oscillatory motion of the particle is a straight line such that its displacement from the equilibrium position (or mean position) varies sinusoidally (harmonically) with time. Since this is the simplest type of periodic motion (or harmonic motion), it is called simple harmonic motion.

Analysis of S.H.M.

Consider a particle (see Fig. 25.1 (i)) executing simple harmonic motion about point O (equilibrium position) with an amplitude (i.e., maximum displacement from O).

Fig. 25.1

The particle is said to have completed 1 vibration (or oscillation) if starting from O, it has moved through positions O → P → O → P′ → O, i.e., when it has returned to its starting position and is moving in the same direction. The time taken to complete 1 vibration is called the period (T) of the motion. The number of complete oscillations in one second is called the frequency (f) of the motion. Clearly, f = 1/T.

If the motion is taken to start from the equilibrium position (i.e., x = 0 at t = 0), then the displacement (x) of the particle from the equilibrium position varies with time (t) according to the relation:

x = a sin ωt = a sin (2πt / T)

where a and ω (or T) are constants of the motion. In order to give physical significance to these constants, it is convenient to plot x–t (displacement–time) graph as in Fig. 25.1 (ii).

The constant a is called the amplitude of the simple harmonic motion; it is the maximum displacement of the particle from the equilibrium position in either direction.
The constant T is called the time period of the simple harmonic motion and is the time taken to complete one oscillation.
The constant f (= 1/T) is called the frequency of oscillation; it is the number of oscillations completed in one second.
The constant ω (= 2π/T) is called the angular frequency of oscillation. It gives the rate of change of phase angle with time.

Let us interpret equation x = a sin ωt in terms of the initial conditions.

When t = 0, x = 0, i.e., the particle is at the equilibrium position.
When t = T/4, x = a, i.e., the particle is at the extreme position.
When t = T/2, x = 0 again, i.e., back at equilibrium.
When t = 3T/4, x = −a, i.e., particle at the negative extreme position.
When t = T, x = 0, i.e., motion is completed and repeated.

25.3 — Periodic Functions

25.3. Periodic Functions

Periodic functions are those which represent periodic motion. The sines and cosines functions of time are simple periodic functions. A function is said to be periodic if it repeats itself after time period T. The same function is obtained when the variable is changed to t + T. Consider the following periodic functions:

f(t) = sin( 2πt / T ) g(t) = cos( 2πt / T )

Here T is the time period of the periodic motion and is equal to 2π radians. We shall see that if the variable t is changed to t + T, the same function results.

f(t + T) = sin [ (2π/T)(t + T) ] = sin ( 2πt/T + 2π ) = sin ( 2πt/T ) = f(t)

Similarly,

g(t + T) = cos ( 2π(t+T)/T ) = cos ( 2πt/T + 2π ) = cos ( 2πt/T ) = g(t)

It can be easily verified that:

f(t + nT) = f(t) g(t + nT) = g(t) (where n = 1, 2, 3, …)

Therefore, infinite sets of periodic function of period T may be represented as:

f_n(t) = sin( 2nπt / T ) g_n(t) = cos( 2nπt / T )

Further, a linear combination of sine and cosine functions like:

y = A sin ωt + B cos ωt …(i)

is also a periodic function with a period T = 2π/ω.

Eq. (i) can be written as: y = C sin( ωt + α )

where
C = √(A² + B²) , tan α = (B/A)

since sin(ωt + α) = sin ωt cos α + cos ωt sin α.

25.5 — General Displacement Equation of S.H.M.

25.5. General Displacement Equation of S.H.M.

The displacement of S.H.M., y = a sin ωt, represents the special case when the motion is taken to start from the equilibrium (mean) position (i.e., at t = 0, y = 0). In practice the clock may start when the oscillator is at any point of its cycle. If the motion starts with some initial phase, we write the general displacement equation:

x = a · sin( ωt + φ )

Here a is the amplitude, ω the angular frequency, and φ the phase constant (also called initial phase or epoch). If the instant the particle is at the mean position is taken as origin of time, then φ = 0 and x = a sin ωt. Using cosine instead of sine gives an equivalent form (a shift of π/2 in phase):

x = a · cos( ωt + θ )

Either sine or cosine may be used throughout; once chosen, be consistent through the problem.

Fig. 25.3 Reference circle for S.H.M.; projection of OP on AA′ gives the displacement x.

If the point on the reference circle is at angle ωt + φ from the reference line, the projection on the vertical diameter AA′ is

OM = a · sin( ωt + φ ) ⇒ x = a · sin( ωt + φ )

Differentiating the representation or shifting the phase by π/2 shows that the cosine form

x = a · cos( ωt + φ )

is entirely equivalent; it simply corresponds to choosing a different reference for measuring phase.

Notes.

When t = 0, x = a sin φ (or a cos φ), the particle is generally not at the mean position.
At the extremes: x = ±a when ωt + φ = (2n+1)π/2.
At the mean position: x = 0 when ωt + φ = nπ.

Example 25.1. A simple harmonic oscillation is represented by x = 3.4 · cos ( 300 t + 0.74 ) (x in cm, t in s) Determine: (i) amplitude, (ii) frequency and time period, (iii) phase, and (iv) initial phase (epoch).

Solution.

(i) Amplitude: a = 3.4 cm.
(ii) Angular frequency: ω = 300 rad s⁻¹. Frequency: f = ω/(2π) ≈ 47.75 Hz. Period: T = 1/f ≈ 0.02094 s.
(iii) The instantaneous phase is ωt + 0.74.
(iv) Initial phase (at t = 0): 0.74 rad. Initial displacement: x(0) = a cos 0.74 ≈ 2.51 cm toward the positive side.

Worked Relations & Examples — S.H.M.

Relations for S.H.M. used in this section

For a particle in simple harmonic motion, the displacement and velocity may be written as

x = a cos(ωt + θ) …(i) v = dx/dt = −a ω sin(ωt + θ) …(ii)

Squaring and adding (i) and (ii) gives the identity used repeatedly:

(x/a)² + (v/(aω))² = cos²(ωt+θ) + sin²(ωt+θ) = 1. …(iii)

Any initial conditions at t = 0 (for example, known values of x and v) can be inserted into (i)–(ii) to find the phase constant θ and the amplitude a.

Example 25.7.

The period of a particle executing S.H.M. is T = 2 s and it can go to and fro from the equilibrium position at a maximum distance of 5 cm. If at the start of the motion, the particle is in the position of maximum displacement towards the right of the equilibrium position, then write the displacement equation of the particle.

Solution. The general equation for S.H.M. is

y = a sin(ωt + φ) .

Given T = 2 s, so ω = 2π/T = π rad s⁻¹.
Amplitude a = 5 cm.
“Starts at maximum displacement to the right” ⇒ at t = 0, y = +a. In the sine form this requires sin φ = 1 ⇒ φ = π/2.

Hence the displacement is

y = 5 sin(π t + π/2) = 5 cos(π t) (y in cm, t in s).

Example 25.8.

In what time after its motion begins, will a particle oscillating according to the equation y = 7 sin(0.5π t) move from the mean position to maximum displacement?

Solution.

y = 7 sin(0.5π t) …(i)

The standard S.H.M. form is y = a sin(ωt). Comparing with (i),

a = 7 cm, ω = 0.5π rad s⁻¹.

Time from mean position (y = 0) to maximum (y = a) corresponds to sin(ωt) = 1 ⇒ ωt = π/2.

t = (π/2) / ω = (π/2) / (0.5π) = 1 s.

Answer: 1 s.

S.H.M. Energy and Example

Energy in S.H.M.

This can be seen by putting v = 0 and y = 0 in eq. (i) and eq. (ii) respectively. In other positions, both forms of energy exist.

Graphical representation. Fig. 25.6 shows the graph of kinetic energy and potential energy versus displacement. Each curve is a parabola centered at y = 0.

E_p = ½ m ω² y² ; E_k = ½ m ω² (a² − y²)

When the particle is at the mean position, y = 0 so that E_p = 0 and E_k = ½ m ω² a². In this position, the total energy of the particle is in the form of kinetic energy since E_p = 0. This is quite clear from Fig. 25.6. Since the kinetic energy of the particle passing through mean position is maximum, it is clear that at this position, the speed of the particle will be maximum.

When the particle is at the extreme position y = a,

E_p = ½ m ω² a² and E_k = 0.

In this position, the total energy of the particle is in the form of potential energy since E_k = 0. This is also clear from Fig. 25.6. Since the kinetic energy of the particle at the extreme positions is zero, the speed of the particle at these positions will be zero.

It is clear from Fig. 25.6 that energy is being continuously transferred between potential energy and kinetic energy of the particle. Note that both E_p and E_k are always positive and their sum at all times is a constant equal to ½ m ω² a². This is represented by a straight line parallel to the displacement axis.

Fig. 25.6: Energy distribution in S.H.M.

Example 25.13.

A mass of 0.5 kg connected to a light spring of force constant 20 N m⁻¹ oscillates on a horizontal frictionless surface.

(i) Calculate the total energy of the system and the maximum speed of the mass if the amplitude of the motion is 3 cm.
(ii) What is the velocity of the mass when the displacement is equal to 2 cm?
(iii) Find the kinetic and potential energies of the system when the displacement equals 2 cm.

Solution.

(i) The total energy of the system is equal to the initial energy stored in the spring for amplitude a = 3 cm = 3 × 10⁻² m.

E = ½ k a² = ½ × 20 × (3 × 10⁻²)² = 9.0 × 10⁻³ J

When the mass is at x = 0 (equilibrium position), E_p = 0 and E_k has the maximum value.

V_max = √(2E / m) = √(2 × 9.0 × 10⁻³ / 0.5) = 0.19 m s⁻¹

(ii) When displacement x = 2 cm,

Total energy, E = E_k + E_p
= ½ m v² + ½ k x²

SHM Energy Problems

Example 25.15.

A particle executes S.H.M. of amplitude a. At what distance from the mean position its K.E. is equal to its P.E.?

Solution.

K.E. = ½ m ω² (a² − y²) ; P.E. = ½ m ω² y²

At mean position K.E. = P.E.

a² − y² = y²
a² = 2y²
y = a / √2 = 0.707 a

Therefore, the distance from mean position where K.E. = P.E. is 0.707 a.

Example 25.16.

A body of mass 1.0 kg is executing S.H.M. given by :

x = 6.0 cos (100π t + π/4) cm

Determine the maximum kinetic energy of the body.

Given:

x = 6.0 cos (100π t + π/4) cm

Comparing it with standard equation of S.H.M. x = a cos(ωt + φ),

a = 6.0 cm = 0.06 m ; ω = 100π = 100 × 3.14 ≈ 314 rad/s

Maximum K.E. = ½ m (aω)²

= ½ × 1.0 × (0.06 × 314)²
= 18 J

PROBLEMS FOR PRACTICE

A particle of mass 0.25 kg vibrates with a period of 2.0 s. If its greatest displacement is 0.04 m, what is its maximum kinetic energy? [0.2 J]
A spring with k = 1.6 × 10³ N m⁻¹ force constant and a mass 0.10 kg attached to it has total vibrational energy of 0.32 J. (i) Calculate its amplitude of vibration (ii) Calculate its maximum speed. [0.02 m ; 8.0 m s⁻¹]
A 3.0 kg ball attached to a spring of negligible mass has force constant of 2.0 × 10³ N m⁻¹. The ball is displaced 0.10 m from equilibrium and then released from rest. What is (i) the maximum speed of the ball (ii) its speed when displacement is 0.05 m? [8.2 m s⁻¹ ; 7.1 m s⁻¹]
Two joules are required to extend a spring by 0.25 m. What is the spring constant? [64 N m⁻¹]
The force constant of a weightless spring is 16 N m⁻¹. A body of mass 1.0 kg is suspended from it. The body is pulled down through 5 cm and then released. (i) Find period of oscillation (ii) Find maximum K.E. of the mass. [(1.57 s) ; (0.2 J)]

25.9 DYNAMICS OF SIMPLE HARMONIC MOTION

We have defined simple harmonic motion (S.H.M.) but we have not described its cause. Let us investigate what causes S.H.M. Consider a particle of mass m executing S.H.M. with amplitude a and angular frequency ω. Then acceleration is

a = − ω² y

Here displacement y is measured from mean position. The negative sign shows that acceleration is always directed towards the mean position. The particle is subjected ...

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Oscillations

INTRODUCTION

25.1 Periodic and Oscillatory Motion

1. Periodic motion

2. Oscillatory motion

25.2. Simple Harmonic Motion (S.H.M.)

Analysis of S.H.M.

25.3. Periodic Functions

25.5. General Displacement Equation of S.H.M.

Relations for S.H.M. used in this section

Example 25.7.

Example 25.8.

Energy in S.H.M.

Example 25.13.

Example 25.15.

Example 25.16.

PROBLEMS FOR PRACTICE

25.9 DYNAMICS OF SIMPLE HARMONIC MOTION

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