|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 potential energy (U) of a particleexecuting simple harmonic motion is,
U(x) = ½ kx2
U= ½ k A2 cos2 (ωt + φ)
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,
= (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.