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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 mω^{2}(a^{2 }– x^{2}) =1/2 x 10 x 1^{2}(10^{2 }– 8^{2})

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

Potential energy = 1/2 mω^{2}x^{2}

∴ Potential energy = 1/2 x 10 x 1^{2 }x 8^{2} = 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 mω^{2}(a^{2 }– x^{2}) =1/2 x 10 x π^{2}(5^{2 }– 2.5^{2})

∴ Kinetic energy = 5 x 3.142^{2}(25^{ }– 6.25) = 5 x 3.142^{2}(18.75)

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

Potential energy = 1/2 mω^{2}x^{2}

∴ Potential energy = 1/2 x 10 x π^{2 }x 2.5^{2} = 5 x 3.142^{2 }x 2.5^{2}

= 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 = v_{max}= ?

The speed when crossing mean position is a maximum speed

Total energy = 1/2 mω^{2}a^{2}

∴ 25 = 1/2 x 0.5 x ω^{2}a^{2}

∴ ω^{2}a^{2} = 25 x 2/ 0.5 = 100

∴ ωa = 10 m/s

but ωa = v_{max }= 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 mω^{2}x^{2 }= 1/2 mω^{2}(a^{2 }– x^{2})

∴ x^{2 }= a^{2 }– x^{2}

∴ 2x^{2} = a^{2}

∴ x = ± a/√2 = ±10/√2 = ±5√2 cm

**Ans:** At a distance of 5√2 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 mω^{2}(a^{2 }– x^{2}) = 3 x 1/2 mω^{2}x^{2 }

∴ a^{2 }– x^{2 }= 3x^{2 }

∴ 4x^{2} = a^{2}

∴ 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 mω^{2}x^{2 }= 1/2 mω^{2}(a^{2 }– x^{2})

∴ x^{2 }= a^{2 }– x^{2}

∴ 2x^{2} = a^{2}

∴ 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

∴ 5x^{2} = 4a^{2} – 4x^{2}

∴ 9x^{2} = 4a^{2}

∴ 4a^{2} = 9x 4^{2} = 144

∴ a^{2} = 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 10^{7}= 3 erg, Force = 2.25 x 10^{-5}N = 2.25 x 10^{-5}x 10^{5 }= 2.25 dyne**To Find:**Distance = x =?

T.E =1/2 mω^{2}a^{2}

∴ 3 =1/2 mω^{2}(2)^{2}

∴ mω^{2} =3/2 ………… (1)

Now Force F = mf = mω^{2}x

∴ 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 = mω^{2}x

F = 100 x (π/2)^{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)^{2}x 10^{2 }= 1.234 x 10^{4} erg

T.E. = 1.234 x 10^{4} 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 mω^{2}(a^{2 }– x^{2}) =1/2 x 200 x 2^{2}(10^{2 }– 3^{2})

∴ Kinetic energy = 100 x 4 x (100 -9) = 3.64 x 10^{4} erg

∴ Kinetic energy = 3.64 x 10^{4} x 10^{-7} J = 3.64 x 10^{-3} J

Potential energy = 1/2 mω^{2}x^{2}

∴ Potential energy = 1/2 x 200 x 2^{2 }x 3^{2} = 3.6 x 10^{3} J

∴ Potential energy = = 3.6 x 10^{3} 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

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