Class 11 Physics Oscillations Energy in SHM

Energy in SHM

The Kinetic and Potential energies in a SHM varies between 0 and their maximum values.

Kinetic energy, potential energy and the total energy is a function of time in the above graph. BothKinetic energy and potential energy repeats after time T/2.

Kinetic energy, potential energy and the total energy is a function of displacement in the above graph.

The kinetic energy (K) of a particle executing SHM can be defined as

K= ½ mv2

= ½ mω2A2sin2 (ωt + φ)

K=½ k A2 sin2 (ωt + φ)

  • The above expression is a periodic function of time, being zero when the displacement is maximum and maximum when the particle is at the mean position.

The potential energy (U) of a particleexecuting simple harmonic motion is,

U(x) = ½ kx2

U= ½ k A2 cos2 (ωt + φ)

  • The potential energy of a particle executing simple harmonic motion is alsoperiodic, with period T/2, being zero at the mean position and maximum at the extremedisplacements.

Total energy of the system always remains the same

E = U + K

= ½ k A2 sin2 (ωt + φ) + ½ k A2 cos2 (ωt + φ)

E=½ k A2(sin2 (ωt + φ) + cos2 (ωt + φ))


The above expression can be written as

E = ½ k A2

Total energy is always constant.

Problem: -A block whose mass is 1 kgis fastened to a spring. The spring has a

spring constant of 50 N m–1. The block ispulled to a distance x = 10 cm from its equilibrium position at x = 0 on a frictionlesssurface from rest at t = 0. Calculate thekinetic, potential and total energies of theblock when it is 5 cm away from mean position?

Answer: -The block executes SHM, its angularfrequency, according to equation, ω= √k/m

= √ (50 N m–1)/ 1kg

= 7.07 rad s–1

Its displacement at any time t is then given by,

x (t) = 0.1 cos (7.07t)

Therefore, when the particle is 5 cm away fromthe mean position, we have

0.05 = 0.1 cos (7.07t)

Or cos (7.07t) = 0.5 and hence

sin (7.07t) = √3/2= 0.866

Then, the velocity of the block at x = 5 cm is

 = 0.1 7.07 0.866 m s–1

 = 0.61 m s–1

Hence the K.E. of the block,

=1/2 m v2

 = ­[1kg (0.6123 m s–1 )2 ]

 = 0.19 J

The P.E. of the block,

=1/2 kx2

 = ­ (50 N m–1 0.05 m 0.05 m)

 = 0.0625 J


The total energy of the block at x = 5 cm,

 = K.E. + P.E.

 = 0.25 J

we also know thatmaximum displacement,K.E. is zero and hence the total energy of thesystem is equal to the P.E. Therefore, the totalenergy of the system,

 = ­ (50 N m–1 0.1 m 0.1 m)

 = 0.25 J , which is same as the sum of the two energies ata displacement of 5 cm.This shows the result with the accordance of conservation of energy.

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