- Periodic Motion
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- Simple Harmonic Motion
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- Equation of SHM
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- Characterstics of SHM
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- Velocity of SHM
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- Acceleration of SHM
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- Total Energy of SHM
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- Motion of a body suspended from a spring
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- Simple pendulum
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- The compound pendulum
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- Damped Oscillations
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- Driven or Forced Harmonic oscillator

- 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.

- 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)

Class 11 Maths page Class 11 Physics page

- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- NCERT Exemplar Problems: Solutions Physics Class 11
- H.C. Verma Concepts of Physics - Vol. 1
- CBSE All in One Physics Class 11 by Arihant
- NCERT Solutions: Physics Class 11th
- New Simplified Physics: A Reference Book for Class 11 (Set of 2 Parts)
- Pradeep's A Text Book of Physics with Value Based Questions - Class XI (Set of 2 Volumes)
- Oswaal CBSE Sample Question Papers For Class 11 Physics (For 2016 Exams)