Plane polarized oscilation of a spring

[DaTario]

Hi All,

Suppose you have a spring with a given elastic constant k. Its natural width is L = 1 meter and its mass is M = 0.1 kg. Now imagine you put this spring to oscilate, with its ends fixed, forming a pure second harmonic oscilation, plane polarized, with frequency Omega. The amplitude of this oscillation is A = 1 meter also.
How is one to determine the energy E of this system ? (mechanical energy of this system E = Ekin + Epot)

P.S. it seems natural that the frequency Omega is to be determined from the constants I alluded above.

P.S. 2: by second harmonic I mean a sine function between the end points (x = 0 and x = 1) with just one crest and one valley.

Best Regards

DaTario

 

[DaTario]

Wait, be faithful, may be the help is to arrive soon...

Best Regards

DaTario

 

[kyle8921]

Hello DaTario,

Thank you for your question! I would approach this problem by first considering the basic principles of energy in a system. In this case, the total energy of the system can be calculated as the sum of its kinetic energy (E_kin) and potential energy (E_pot).

To determine the kinetic energy, we need to know the mass of the spring and its velocity. Since the spring is undergoing a second harmonic oscillation, its velocity can be expressed as v = A*Omega*cos(Omega*t), where A is the amplitude and Omega is the frequency. Therefore, the kinetic energy can be calculated as E_kin = 0.5*M*(A*Omega*cos(Omega*t))^2.

Next, we need to determine the potential energy of the system. In this case, the potential energy is due to the elastic potential energy stored in the spring. This can be expressed as E_pot = 0.5*k*x^2, where k is the elastic constant and x is the displacement of the spring from its equilibrium position. Since the spring is undergoing a second harmonic oscillation, the displacement can be expressed as x = A*sin(Omega*t). Therefore, the potential energy can be calculated as E_pot = 0.5*k*(A*sin(Omega*t))^2.

Finally, we can calculate the total energy of the system by summing the kinetic and potential energies: E = E_kin + E_pot. By substituting the expressions we derived above, we get E = 0.5*M*(A*Omega*cos(Omega*t))^2 + 0.5*k*(A*sin(Omega*t))^2. This expression gives us the total energy of the system at any given time during the oscillation.

To find the frequency Omega of the oscillation, we can use the relationship between frequency, mass, and spring constant, which is given by Omega = sqrt(k/M). Therefore, by plugging in the values of k and M given in the problem, we can determine the frequency of the oscillation.

I hope this explanation helps you understand how to determine the energy of a system undergoing plane polarized oscillation of a spring. Let me know if you have any further questions. Best regards,