Numerical Problems on Energy of S.H.M.

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Example – 1:

  • A particle of mass 10 g performs S.H. M. of amplitude 10 cm and period 2π s. Determine its kinetic and potential energies when it is at a distance of 8 cm from its equilibrium position.
  • Solution:
  • Given: Mass = m = 10 g, amplitude = a = 10 cm, Period = T = 2π s, displacement = x = 8 cm
  • To Find: Kinetic energy = ? and Potential energy = ?

Angular velocity ω = 2π/T =  2π/2π = 1 rad/s

Kinetic energy = 1/2 mω2(a– x2) =1/2 x 10 x 12(10– 82)

∴  Kinetic energy  = 5 x (36) = 180 erg = 1.8 x 10-5 J

Potential energy = 1/2 mω2x2



∴  Potential energy  = 1/2 x 10 x 1x 82 = 320 erg = 3.2 x 10-5 J

Ans: Kinetic energy = 1.8 x 10-5 J and potential energy =  3.2 x 10-5 J

Example – 2:

  • A particle of mass 10 g executes linear S.H.M. of amplitude 5 cm with a period of 2 s. Find its PE and KE, 1/6 s after it has crossed the mean position.
  • Solution:
  • Given: Mass = m = 10 g, amplitude = a = 5 cm, Period = T = 2 s, time elapsed = 1/6 s, particle passes through mean position, α = 0.
  • To Find: Kinetic energy =? and Potential energy =?

Angular velocity ω = 2π/T =  2π/2 = π rad/s

Displacement of a particle performing S.H.M. is given by



x = a sin (ωt + α)

∴  x = 5 sin ( π x 1/6 + 0)

∴  x = x = 5 sin ( π/6) = 5 x 1/2 = 2.5 cm

Kinetic energy = 1/2 mω2(a– x2) =1/2 x 10 x π2(5– 2.52)

∴  Kinetic energy = 5 x 3.1422(25 – 6.25) =  5 x 3.1422(18.75)



∴  Kinetic energy = 925.5 erg = 9.26 x 10-5 J

Potential energy = 1/2 mω2x2

∴  Potential energy  = 1/2 x 10 x πx 2.52 = 5 x 3.142x 2.52

= 308.5 erg = 3.09 x 10-5 J

Ans: Kinetic energy = 9.26 x 10-5 J and potential energy =  3.09 x 10-5 J



Example – 3:

  • The total energy of a particle of mass 0.5 kg performing S.H.M. is 25 J. What is its speed when crossing the centre of its path?
  • Solution:
  • Given: Mass = m = 0.5 kg, Total energy T.E. = 25 J
  • To Find: Maximum speed = vmax = ?

The speed when crossing mean position is a maximum speed

Total energy = 1/2 mω2a2

∴  25 = 1/2 x 0.5 x ω2a2

∴  ω2a2 = 25 x 2/ 0.5 = 100

∴  ωa = 10 m/s



but ωa = vmax = 10 m/s

Ans: The speed when crossing mean position is 10m/s

Example – 4:

  • A particle performs a linear S.H.M. of amplitude 10 cm. Find at what distance from the mean position its PE is equal to its KE.
  • Solution:
  • Given: P.E. = K.E.
  • To Find: Distance = x=?

P.E. = K.E.

∴  1/2 mω2x= 1/2 mω2(a– x2)



∴ x2  =  a– x2

∴ 2x2 = a2

∴ x = ± a/√2 = ±10/√2  = ±52  cm

Ans:  At a distance of 52  cm fon either side of the mean position K.E. = P.E.

Example – 5:

  • Find the relation between amplitude and displacement at the instant when the K.E. of a particle performing S.H. M. is three times its P.E.
  • Solution:
  • Given: K.E. = 3 x P.E.
  • To Find: Distance = x=?

K.E. = 3 x P.E.

∴  1/2 mω2(a– x2)  = 3 x 1/2 mω2x



∴ a– x= 3x

∴ 4x2 = a2

∴ x = ± a/2 , where a = amplitude

Ans:  At a distance of a/2  cm fon either side of the mean position K.E. = 3 x P.E.



Example – 6:

  • When is the displacement in S.H.M. one-third of the amplitude, what fraction of total energy is kinetic and what fraction is potential? At what displacement is the energy half kinetic and half potential?
  • Solution:
  • Part – I:
  • Given: x = a/3
  • To Find: K.E/T.E. =? and P.E./T.E. =?

  • Part – II
  • Given: P.E. = K.E.
  • To Find: Distance = x = ?

P.E. = K.E.

∴  1/2 mω2x= 1/2 mω2(a– x2)

∴ x2  =  a– x2

∴ 2x2 = a2



∴ x = ± a/√2

Ans:  The fraction of K.E = 8/9, fraction of P.E. = 1/9,  required displacement = ± a/√2 unit

Example – 7:

  • An object of mass 0.2 kg executes S.H.M. along the X-axis with a frequency of 25 Hz. At the position x = 0.04 m, the object has a K.E. of 0.5 J and P.E. of 0.4 J. Find the amplitude of its oscillations
  • Solution:
  • Given: Mass = m = 0.2 kg, frequency = n = 25 Hz, displacement = x = 0.04 m = 4 cm, K.E. = 0.5 J, P.E. = 0.4 J
  • To Find: Amplitude = a = ?

Angular speed ω = 2πn = 2 x π x 25 = 50π rad/s

Kinetic energy

∴   5x2 = 4a2 – 4x2

∴   9x2 = 4a2

∴ 4a2 =  9x 42 = 144

∴ a2 =  36

∴  a = 6 cm

Ans: The amplitude = 6 cm

Example – 8:

  • The amplitude of a particle in S.H. M. is 2 cm and the total energy of its oscillation is 3 x 10-7 J. At what distance from the mean position will the particle be acted upon by a force of 2.25 x 10-5 N when vibrating?
  • Solution:
  • Given: amplitude = a = 2 cm, Total energy = T.E. = 3 x10-7 J = 3 x10-7 x 107 = 3 erg, Force = 2.25 x 10-5 N =  2.25 x 10-5 x 10= 2.25 dyne
  • To Find: Distance = x =?

T.E =1/2 mω2a2

∴ 3 =1/2 mω2(2)2

∴ mω2 =3/2 ………… (1)

Now Force F = mf = mω2x

∴  2.25 = (3/2)x

∴ x = 2.25 x 2 /3 = 1.5 cm

Ans: At a distance of 1.5 cm from the mean position will the particle be acted upon by a force of 2.25 x 10-5 N

Example – 9:

  • A body of mass 100 g performs S.H.M. along a path of length 20 cm and with a period of 4 s. Find the restoring force acting upon it at a displacement of 3 cm from the mean position? Find also the total energy of the body.
  • Solution:
  • Given: mass = m = 20 g, Path length = 20 cm, amplitude = a = 20/2 = 10 cm, Period = T = 4s,
  • To Find: Restoring force = F = ?, Total energy = T.E. = ?

Angular speed ω = 2π/T = 2π/4  = π/2 rad/s

Restoring force F = mf = mω2x

F =  100 x (π/2)x 3 = 740.4 dyne = 740.4 x 10-5 N = 7.404 x 10-3 N

T.E. = 1/2 x 100 x (π/2)2x 10= 1.234 x 104 erg

T.E. =  1.234 x 104 x 10-7 J =  1.234 x 10-3 J

Ans: Restoring force = 7.404 x 10-3 N; total energy = 1.234 x 10-3 J

Example – 10:

  • A particle of mass 200 g performs S.H.M. of amplitude 0.1m and period 3.14 second. Find its K.E. and P.E. when it is at a distance of 0.03 m from the mean position.
  • Solution:
  • Given: mass = m = 200 g, amplitude = a = 0.1 m = 10 cm, period = T = 3.14 s, Distance = x = 0.03 m = 3 cm,
  • To Find: K.E. =? and P.E. = ?

Angular speed ω = 2π/T = 2π/3.14  = 2 rad/s

Kinetic energy = 1/2 mω2(a– x2) =1/2 x 200 x 22(10– 32)

∴  Kinetic energy = 100 x 4 x (100 -9) =  3.64 x 104 erg

∴  Kinetic energy = 3.64 x 104 x 10-7 J =  3.64 x 10-3 J

Potential energy = 1/2 mω2x2

∴  Potential energy  = 1/2 x 200 x 2x 32 = 3.6 x 103 J

∴  Potential energy  = = 3.6 x 103 x 10-7 = 3.6 x 10-4 J

Ans: K.E. = 3.64 x 10-3 J; P.E. = 3.6 x 10-4 J

Science > Physics > OscillationsYou are Here
Physics Chemistry  Biology  Mathematics

2 Comments

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