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Energy in Simple Harmonic Motion

Simple Harmonic Motion

The total energy (E) of an oscillating particle is equal to the sum of its kinetic energy and potential energy if conservative force acts on it.

The velocity of a particle executing SHM at a position where its displacement is y from its mean position is v = ω√a2 – y2

Kinetic energy

Kinetic energy of the particle of mass m is,

K = ½ m [ω√a2 – y2]2

          …... (1)

Potential energy

From definition of SHM F = –ky the work done by the force during the small displacement dy is dW = −F.dy = −(−ky) dy = ky dy

∴ Total work done for the displacement y is,

W = \int dW = \int_{0}^{y}ky dy

As k = mω2, therefore,

W =\int_{0}^{y}m\omega ^{2}y dy 

Thus, W = ½ mω2y2                …... (2)

This work done is stored in the body as potential energy.

        …... (3)

Total energy, E = K+U

= ½ mω(a2 – y2) + ½ mω2y2

= ½ mω2a2


          …... (4)

Thus we find that the total energy of a particle executing simple harmonic motion is ½ mω2a2.

Special cases

Energy - Displacement Graph

(i) When the particle is at the mean position y = 0, from eqn (1) it is known that kinetic energy is maximum and from eqn. (2) it is known that potential energy is zero. Hence the total energy is wholly kinetic.

E = Kmax = ½ mω2a2

(ii)  When the particle is at the extreme position y = +a, from equation (1), it is known that kinetic energy is zero and from eqn. (2) it is known that Potential energy is maximum. Hence the total energy is wholly potential.

E = Umax = ½ mω2a2

(iii) When y = a/2,

K = ½ mω2[a2 – a2/4]

So, K = ¾ (½ mω2a2)

K = ¾ E

U = ½ mω2(a/2)2 = ¼ (½ mω2a2)

So, U = ¼ E

If the displacement is half of the amplitude, K = ¾ E and U = ¼ E. K and U are in the ratio 3 : 1.

E = K+U = ½ mω2a2

At any other position the energy is partly kinetic and partly potential.

This shows that the particle executing SHM obeys the law of conservation of energy.

Refer this video to know more about, “Energy in Simple Harmonic Motion”.

Graphical representation of energy

The values of K and U in terms of E for different values of y are given in the beloe table. The variation of energy of an oscillating particle with the displacement can be represented in a graph as shown in the figure.

Types of oscillations

There are three main types of oscillations.

(i) Free oscillations

When a body vibrates with its own natural frequency, it is said to execute free oscillations. The frequency of oscillations depends on the inertial factor and spring factor, which is given by,

n = 1/2π √k/m


(a) Vibrations of tuning fork

(b) Vibrations in a stretched string 

(c) Oscillations of simple pendulum

(d) Air blown gently across the mouth of a bottle.

(ii) Damped oscillations

Damped Oscillations

Most of the oscillations in air or in any medium are damped. When an oscillation occurs, some kind of damping force may arise due to friction or air resistance offered by the medium. So, a part of the energy is dissipated in overcoming the resistive force. Consequently, the amplitude of oscillation decreases with time and finally becomes zero. Such oscillations are called damped oscillations as shown in figure.

Examples :

(a) The oscillations of a pendulum

(b) Electromagnetic damping in galvanometer (oscillations of a coil in galvanometer)

(c) Electromagnetic oscillations in tank circuit

(iii) Maintained oscillations

Maintained Oscillations

The amplitude of an oscillating system can be made constant by feeding some energy to the system. If an energy is fed to the system to compensate the energy it has lost, the amplitude will be a constant. Such oscillations are called maintained oscillations as shown in figure.

Example :

A swing to which energy is fed continuously to maintain amplitude of oscillation.

(iv) Forced oscillations

When a vibrating body is maintained in the state of vibration by a periodic force of frequency (n) other than its natural frequency of the body, the vibrations are called forced vibrations. The external force is driver and body is driven.

The body is forced to vibrate with an external periodic force. The amplitude of forced vibration is determined by the difference between the frequencies of the driver and the driven. The larger the frequency difference, smaller will be the amplitude of the forced oscillations.

Examples :

Sound boards of stringed instruments execute forced vibration,

Press the stem of vibrating tuning fork, against tabla. The tabla suffers forced vibration.



In the case of forced vibration, if the frequency difference is small, the amplitude will be large as shown in figure. Ultimately when the two frequencies are same, amplitude becomes maximum.This is a special case of forced vibration.

If the frequency of the external periodic force is equal to the natural frequency of oscillation of the system, then the amplitude of oscillation will be large and this is known as resonance.


(i) Using resonance, frequency of a given tuning fork is determined with a sonometer.

(ii) In radio and television, using tank circuit, required frequency can be obtained.


(a) Resonance can cause disaster in an earthquake, if the natural frequency of the building matches the frequency of the periodic oscillations present in the Earth. The building begins to oscillate with large amplitude thus leading to a collapse.

(b) A singer maintaining a note at a resonant frequency of a glass, can cause it to shatter into pieces.

  • The sum of the kinetic and potential energies in a simple harmonic oscillator is a constant, i.e., KE+PE=constant. The energy oscillates back and forth between kinetic and potential, going completely from one to the other as the system oscillates.

  • The maximum velocity depends on three factors: amplitude, the stiffness factor, and mass:

  • The total energy is a constant of the motion, as expected for an isolated system. Moreover, the energy is proportional to the amplitude squared of the motion.

  • The kinetic energy attains its maximum value, and the potential energy attains it minimum value, when the displacement is zero.

  • Likewise, the potential energy attains its maximum value, and the kinetic energy attains its minimum value, when the displacement is maximal.

  • The minimum value of K is zero, since the system is instantaneously at rest when the displacement is maximal.

Problem (JEE Advanced):

The potential energy of a harmonic oscillator of mass 2 kg in its mean position is 15 J. If its total energy is 9 J and its amplitude is 0.01 m, its time period will be,

(a) (π/100) s                (b) (π/50) s

(c) (π/20) s                  (d) (π/10) s


Here, ½ kA2 = (9 – 5) J = 4 J

So, k = 8/A2 = 8/(0.01)2 = 8\times104 N/m

T = 2π√m/k = 2π√[2/(8\times104)] = (π/100) s

From the above observation, we conclude that, option (a) is correct.

Question 1

A linear harmonic oscillator has a total energy of 160 J. Its

(a) maximum potential energy is 100 J

(b) maximum kinetic energy is 160 J

(c) minimum potential energy is 100 J

(d) maximum kinetic energy is 100 J 

Question 2

A particle executes SHM with an amplitude 4 cm. At what displacement from the mean position its energy is half kinetic and half potential? 

(a) 2√2 cm         (b) √2 cm

(c) 2 cm              (d) 1 cm

Question 3

A force of 6.4 N stretches a vertical spring by 0.1 m. The mass that must be suspended from the spring so that it oscillates with a period of π/4 s is,

(a)  π/4 kg            (b) 1 kg

(c) ¼ kg                (d) 10kg

Question 4

When K.E energy of SHM is maximum its

(a) P.E is zero                            (b) acceleration is zero

(c) restoring force is zero        (d) all P.E acceleration and restoring force are zero

Question 5

In case of a simple pendulum the cause of damping is

(a) drag force of air             (b) gravity

(c)  tension in string            (d) none of these

Q.1 Q.2 Q.3 Q.4 Q.5







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