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प्रश्न
Discuss in detail the energy in simple harmonic motion.
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उत्तर
Energy in simple harmonic motion:
(a) Expression for Potential Energy: For the simple harmonic motion, the force and the displacement are related by Hooke’s law
`vec"F" = -"k"vec("r")`
Since force is a vector quantity, in three dimensions it has three components.
Further, the force in the above equation is a conservative force field; such a force can be derived from a scalar function which has only one component. In one dimensional case
F = − kx .....................(1)
The work done by the conservative force field is independent of the path. The potential energy U can be calculated from the following expression.
F = `-"dU"/"dx"` ..........(2)
Comparing (1) and (2), we get
`-"dU"/"dx" = -"kx"`
dU = kx dx
This work done by the force F during a small displacement dx stores as potential energy
U(x) = `int_0^"x" "kx′" "dx′" = 1/2["k"("x′")^2]_0^"x" = 1/2"kx"^2` ..........(3)
From equation `ω = sqrt("k"/"m")`, we can substitute the value of force constant k = mω2 in equation (3),
U(x) = `1/2"m"ω^2"x"^2` ...................(4)
where ω is the natural frequency of the oscillating system. For the particle executing simple harmonic motion from equation y = A sin ωt, we get
x = A sin ωt
U(t) = `1/2"m"ω^2"A"^2 sin^2ω"t"` ............(5)
This variation of U is shown in figure.
Variation of potential energy with time t
(b) Expression for Kinetic Energy:
Kinetic energy KE = `1/2"mv"_"x"^2 = 1/2"m" ("dx"/"dt")^2` ............(6)
Since the particle is executing simple harmonic motion, from equation y = A sin ωt
x = A sin ωt
Therefore, velocity is
vx = `"dx"/"dt"` = Aω cos ωt ...................(7)
= `"A"ω sqrt(1 - ("x"/"A")^2)`
vx = `ω sqrt("A"^2 - "x"^2)`
Hence, KE = `1/2"mv"_"x"^2 = 1/2"m"ω^2 ("A"^2 - "x"^2)` ........(9)
KE = `1/2"m"ω^2 "A"^2 cos^2 ω"t"` ..............(10)
This variation with time is shown in figure.
Variation of kinetic energy with time t
(c) Expression for Total Energy: Total energy is the sum of kinetic energy and potential energy
E = KE + U ..........…(11)
E = `1/2"m"ω^2 ("A"^2 - "x"^2) + 1/2 "m"ω^2 "x"^2`
Hence, cancelling x2 term,
E = `1/2"m"ω^2"A"^2` = constant ...............(12)
Alternatively, from equation (5) and equation (10), we get the total energy as
E = `1/2"m"ω^2"A"^2 sin^2 ω"t" + 1/2"m"ω^2"A"^2 cos^2 ω"t"`
= `1/2"m"ω^2"A"^2 (sin^2 ω"t" + cos^2 ω"t")`
From trigonometry identity, (sin2 ωt + cos2 ωt) = 1
E = `1/2"m"ω^2"A"^2` = constant
which gives the law of conservation of total energy. This is depicted in Figure
Both kinetic energy and potential energy vary but total energy is constant
Thus the amplitude of a simple harmonic oscillator can be expressed in terms of total energy.
A = `sqrt((2"E")/("m"ω^2)) = sqrt((2"E")/"k")` ..........(13)
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