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Question
Show that for a particle executing simple harmonic motion.
- the average value of kinetic energy is equal to the average value of potential energy.
- 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"`
Solution
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, EK = `1/2 "mv"^2 = 1/2"m"ω^2 "A"^2 cos^2 ω"t"`
Potential energy, EP = `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`
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