Answer to Question #83185 in Classical Mechanics for prince

Question

Calculate the average value of the potential energy over one complete cycle of a simple harmonic oscillator with stiffness k=200 N/m and amplitude a=5 cm
a. 0.125 J
b .1.500 J
c. 0.250 J
d. 0.500J

Answer 1

The instantaneous potential energy of a simple harmonic oscillator

PE(t)=(kx^2 (t))/2

The displacement

x(t)=a cos⁡ωt=a cos^2⁡〖2π/T t〗

The average potential energy

(PE) ̅=1/T ∫_0^T▒PE(t)dt=(ka^2)/2T ∫_0^T▒〖cos^2⁡〖2π/T〗 t〗 dt=

=(ka^2)/4T ∫_0^T▒(1+cos⁡〖4π/T〗 t) dt=(ka^2)/4=(200×〖0.05〗^2)/4=0.125 J