Question

Show that for a particle executing simple harmonic motion.

1. the average value of kinetic energy is equal to the average value of potential energy.
2. average potential energy = average kinetic energy = `1/2` (total energy)

Hint: average kinetic energy = <kinetic energy> = `1/"T" int_0^"T" ("Kinetic energy") "dt"` and

average potential energy = <potential energy> = `1/"T" int_0^"T" ("Potential energy") "dt"`

Answer 1

Suppose a particle of mass m executes SHM of period T.

The displacement of the particles at any instant t is given by y = A sin ωt

Velocity v = `"dy"/"dt" = ω"A" cos ω"t"`

Kinetic energy, E_K = `1/2 "mv"^2 = 1/2"m"ω^2 "A"^2 cos^2 ω"t"`

Potential energy, E_P = `1/2"m"ω^2"y"^2 = 1/2"m"ω^2 "A"^2 sin^2 ω"t"`

a. Average K.E. over a period of oscillation,

`"E"_("K"_"av") = 1/"T" int_0^"T" "E"_"K"  "dt"`

= `1/"T" int_0^"T" 1/2 "m"ω^2 "A"^2 cos^2 ω"t"  "dt"`

= `1/(2"T")"m"ω^2 "A"^2 int_0^"T" [(1 + cos 2  ω"t")/2] "dt"`

= `1/(4"T")"m"ω^2 "A"^2 ["t" + (sin 2  ω"t")/(2ω)]_0^"T"`

= `1/(4"T")"m"ω^2 "A"^2 "T"`

`"E"_("K"_"av") = 1 /4"m"ω^2 "A"^2`

b. Average P.E. over a period of oscillation

`"E"_("P"_"av") = 1/"T" int_0^"T" "E"_"P"  "dt"`

= `1/"T" int_0^"T" 1/2 "m"ω^2 "A"^2 sin^2 ω"t"  "dt"`

= `1/(2"T")"m"ω^2 "A"^2 int_0^"T" [(1 - cos 2  ω"t")/2] "dt"`

= `1/(4"T")"m"ω^2 "A"^2 ["t" - (sin 2  ω"t")/(2ω)]_0^"T"`

= `1/(4"T")"m"ω^2 "A"^2 "T"`

`"E"_("P"_"av") = 1 /4"m"ω^2 "A"^2`