Simple Harmonic Motion
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7. Total energy in SHM
- When a system at rest is displaced from its equilibrium position by doing work on it,
it gains potential energy and when it is released, it begins to move with a velovity and acquires kinetic energy.
- If m is the mass of system executing SHM then kinetic energy of system at any instant of time is
K=(1/2)mv^{2}
(13)
putting equation 8 in 13 we get,
- From equation (14) we see that Kinetic Energy of system varies periodically i.e.,
it is maximum (= (1/2)mω^{2}A^{2}) at the maximum value of velocity
( ±ωA) and at this time displacement is zero.
- When displacement is maximum (±A), velocity of SHM is zero and hence kinetic energy is also zero and
at these extreme points where kinetic energy K=0, all the energy is potential.
- At intermediate positions of lying between 0 and ±A, the energy is partly kinetic and partly potential.
- To calculate potential energy at instant of time consider that x is the displacement of the system from its equilibrium at any time t.
- We know that potential energy of a system is given by the amount of work required to move system
from position 0 to x under the action of applied force.
- Here force applied on the system must be just enough to oppose the restoring force -kx i.e., it should be equal to kx.
- Now work required to give infinitesimal displacement is dx=kx dx.
Thus, total work required to displace the system from 0 to x is
thus,
where, from equation 5 ω=√(k/m) and displacement x=A cos(ωt+φ).
-From equation 14 and 15 we can calculate total energy of SHM which is given by,
- Thus total energy of the oscillator remains constant as displacement is regained after every half cycle.
- If no energy is dissipated then all the potential energy becomes kinetic and vice versa.
- Figure below shows the variation of kinetic energy and potential energy of harmonic oscillator with
time where phase φ is set to zero for simplicity.
8. Some simple systems executing SHM
(A) Motion of a body suspended from a spring
- Figure (6a) below shows a spring of negligible mass, spring constant k and length l suspended from a rigid support.
- When a body of mass m is attached to this spring as shown in figure 6(b), the spring elongates and it would then rest in equilibrium position such that upward force F_{up} exerted by spring is equal to the weight mg of the boby.
- If the spring is extended by an amount Δl ater attachment of block of mass m then in its equilibrium position upward force equals
F_{up}=kΔl
also in this equilibrium position
F_{up}=mg
or, kΔl=mg
- Again the body is displaced in upwards direction such that it is at a distance x above equilibrium position as shown in figure 6(c).
- Now extansion of spring would be (Δl-x), thus upward force now exerted on the body is
F_{up}=k(Δl-x)
- Weight of the body now tends to pull the spring downwards with a force equal to its weight. Thus resultant force on the body is
F=k(Δl-x)-mg
=mg-kx-mg
or,
F=-kx
(17)
- From equation 17 we see that resultant force on the body is proportional to the displacement of the body from its equilibrium position.
- If such a body is set into vertical oscillations it oscillates with an angular frequency
ω=√(k/m) (18)
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