Question

The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5 j. If its total energy is 9 j and its amplitude is 0.01m, its time period would be

Answer 1

Energy in Simple Harmonic Motion



The total energy (E) of an oscillating particle is equal to the sum of its kinetic energy and potential energy if conservative force acts on it.

The velocity of a particle executing SHM at a position where its displacement is y from its mean position is v = ω√a2 – y2

Kinetic energy

Kinetic energy of the particle of mass m is,

K = ½ m [ω√a2 – y2]2

          …... (1)

Potential energy

From definition of SHM F = –ky the work done by the force during the small displacement dy is dW = −F.dy = −(−ky) dy = ky dy

∴ Total work done for the displacement y is,



As k = mω2, therefore,

 

Thus, W = ½ mω2y2                …... (2)

This work done is stored in the body as potential energy.

        …... (3)

Total energy, E = K+U

= ½ mω2 (a2 – y2) + ½ mω2y2

= ½ mω2a2

So,

          …... (4)

Thus we find that the total energy of a particle executing simple harmonic motion is ½ mω2a2.

Special cases



(i) When the particle is at the mean position y = 0, from eqn (1) it is known that kinetic energy is maximum and from eqn. (2) it is known that potential energy is zero. Hence the total energy is wholly kinetic.

E = Kmax = ½ mω2a2

(ii)  When the particle is at the extreme position y = +a, from equation (1), it is known that kinetic energy is zero and from eqn. (2) it is known that Potential energy is maximum. Hence the total energy is wholly potential.

E = Umax = ½ mω2a2

(iii) When y = a/2,

K = ½ mω2[a2 – a2/4]

So, K = ¾ (½ mω2a2)

K = ¾ E

U = ½ mω2(a/2)2 = ¼ (½ mω2a2)

So, U = ¼ E

If the displacement is half of the amplitude, K = ¾ E and U = ¼ E. K and U are in the ratio 3 : 1.

E = K+U = ½ mω2a2

At any other position the energy is partly kinetic and partly potential.

This shows that the particle executing SHM obeys the law of conservation of energy.

Refer this video to know more about, “Energy in Simple Harmonic Motion”.

Answer 2

Maximum velocity occurs at mean position in the simple harmonic motion

K.E at the centre= 9-5 = 4J

=>1/2mv^2 = 4 => V(max) =2m/s

In SHM V(max) = AW where A=1 in the given problem =>W=2 =>f=3.14  (f=2*3.14/W)