Question

A simple harmonic oscillator has amplitude a angular velocity ω and mass m. then average energy in one time period will be

Answer 1

Answer on Question #41179, Physics, Mechanics | Kinematics | Dynamics

A simple harmonic oscillator has amplitude A angular velocity $\omega$ and mass m. Then average energy in one time period will be

Solution:

Linear Harmonic oscillator.

Mechanical model: mass m on a spring characterized by a spring constant k.

Elastic restoring force $F = -kx$ is balanced according to Newton's second law

$$F = m a$$ $$m \ddot{x} = - k x$$

The equation of motion

$$\ddot{x} + \omega^2 x = 0$$

where

$$\omega = \sqrt{\frac{k}{m}}$$

Such system oscillates with amplitude A and angular frequency $\omega$.

Consider solution

$$x = A \sin(\omega t + \varphi)$$

The velocity v is equal to

$$v = \dot{x} = A \omega \cos(\omega t + \varphi)$$

and thus the kinetic energy

$$K = \frac{1}{2} m v^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t + \varphi) = K_0 \cos^2(\omega t + \varphi)$$

where the maximum kinetic energy is equal to

$$K_0 = \frac{1}{2} m A^2 \omega^2 = \frac{1}{2} k A^2$$

The potential energy – work done by applied force displacing the system from 0 to x

$$U(x) = \int_0^x k x \, dx = \frac{1}{2} k x^2$$

Substituting x

$$U(x) = \frac{1}{2} k A^2 \sin^2(\omega t + \varphi) = U_0 \sin^2(\omega t + \varphi)$$

where $U_0$ is the maximum potential energy (for $x = A$)

$$U_0 = \frac{1}{2} k A^2$$

The average values over one oscillation period are calculated using the definition

$$\langle f \rangle = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} f(t) \, dt$$

Thus:

$$\langle U \rangle = \frac {\int_ {0} ^ {T} U d t}{\int_ {0} ^ {T} d t} = \frac {\int_ {0} ^ {T} U _ {0} \sin^ {2} (\omega t + \varphi) d t}{T} = \frac {1}{2} U _ {0} = \frac {1}{2} k A ^ {2}$$

and

$$\langle K \rangle = \frac {\int_ {0} ^ {T} K d t}{\int_ {0} ^ {T} d t} = \frac {1}{2} K _ {0} = \frac {1}{2} k A ^ {2}$$

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.

Thus,

$$\langle E \rangle = \langle K \rangle + \langle U \rangle = \frac {1}{2} k A ^ {2} + \frac {1}{2} k A ^ {2} = k A ^ {2} = m A ^ {2} \omega^ {2}$$

Answer. $\langle E\rangle = mA^2\omega^2$

http://www.AssignmentExpert.com/