show that total energy of a particle performing linear S.H.M is constant
Answers
To show thay total energy of a particle performing linear S.H.M is constant:
1. There are two types of energy in a particle executing simple harmonic motion.
Potential energy which is due to the displacement of the particle from the mean position
Kinetic energy which is due to the velocity of the particle
At any instance, the total energy of the particle is equal to the sum of the kinetic energy and the potential energy.
2. Consider a particle of mass m, having linear simple harmonic motion with amplitude a and constant angular frequency ω. Suppose t seconds after starting from the mean position, the displacement of the particle is given by y = a sin ωt
Velocity of the particle at an instant t is v = dy/dt = a ω cos ωt
Accelaration of the particle at an instant t is v = d²y/dt² = -a ω²sin ωt = - ω²y
The negative sign shows that it is directed towards the mean position.
3. We will now calculate potential energy and kinetic energy separately and then find the total energy.
4. Potential energy (PE):
Restoring force F = mass * acceleration = m * (- ω²y) = -ky where k = spring factor of SHM = m ω²
Work done for the additional displacement dy against the restoring force dW = -F dy = - (-ky) dy = k dy
Total work done in displacing the particle from the mean position to the position of displacement y = W = ∫ ky dy within the limits 0 to y = ½ ky²
Potential energy = PE = ½ ky² = ½ m ω²y²= ½ m ω²a²sin²ωt
5.Kinetic Energy (KE):
Kinetic Energy, KE = ½ mv²= ½ m (a ω cos ωt) = ½ m a²ω²cos² ωt
= ½ m a²ω²(1 - sin² ωt)
= ½ m a²ω² (1 – y² / a² )
= ½ m ω²(a²– y² )
6. Total Energy:
Total energy of a particle at an instant t = PE + KE = ½ m ω²y²+ ½ m ω² (a²– y²) = ½ m ω²a²
Where m. a, ω are all constants. Hence, total energy of the system remains constant at all times