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In this article, we shall study to solve numerical problems to calculate potential energy, kinetic energy, and total energy of particle performing S.H.M.
Example – 01:
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.
Given: Mass = m = 10 g, amplitude = a = 10 cm, Period = T = 2π s, displacement = x = 8 cm
To Find: Kinetic energy =? and Potential energy = ?
Solution:
Angular velocity ω = 2π/T = 2π/2π = 1 rad/s
Kinetic energy = 1/2 mω2 (a2 – x2) =1/2 x 10 x 12(102 – 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 12 x 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 – 02:
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.
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 =?
Solution:
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 (a2 – x2) =1/2 x 10 x π2 (52 – 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 π2 x 2.52 = 5 x 3.1422 x 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 – 03:
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?
Given: Mass = m = 0.5 kg, Total energy T.E. = 25 J
To Find: Maximum speed = vmax =?
Solution:
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 – 04:
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.
Given: P.E. = K.E.
To Find: Distance = x=?
Solution:
P.E. = K.E.
∴ 1/2 mω2x2 = 1/2 mω2 (a2 – x2)
∴ x2 = a2 – x2
∴ 2x2 = a2
∴ x = ± a/√2 = ±10/√2 = ±5√2 cm
Ans: At a distance of 5√2 cm from either side of the mean position K.E. = P.E.
Example – 05:
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.
Given: K.E. = 3 x P.E.
To Find: Distance = x=?
Solution:
K.E. = 3 x P.E.
∴ 1/2 mω2 (a2 – x2) = 3 x 1/2 mω2x2
∴ a2 – x2 = 3x2
∴ 4x2 = a2
∴ x = ± a/2, where a = amplitude
Ans: At a distance of a/2 cm from either side of the mean position K.E. = 3 x P.E.
Example – 06:
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?
Part – I:
Given: x = a/3
To Find: K.E/T.E. =? and P.E./T.E. =?
Solution:
Part – II
Given: P.E. = K.E.
To Find: Distance = x =?
Solution:
P.E. = K.E.
∴ 1/2 mω2x2 = 1/2 mω2(a2 – x2)
∴ x2 = a2 – 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 – 07:
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.
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 =?
Solution:
Angular speed ω = 2πn = 2 x π x 25 = 50π rad/s
∴ 5x2 = 4a2 – 4x2
∴ 9x2 = 4a2
∴ 4a2 = 9x 42 = 144
∴ a2 = 36
∴ a = 6 cm
Ans: The amplitude = 6 cm
Example – 08:
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?
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 105 = 2.25 dyne
To Find: Distance = x =?
Solution:
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 – 09:
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.
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. = ?
Solution:
Angular speed ω = 2π/T = 2π/4 = π/2 rad/s
Restoring force F = mf = mω2x
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)2x 102 = 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.
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. = ?
Solution:
Angular speed ω = 2π/T = 2π/3.14 = 2 rad/s
Kinetic energy = 1/2 mω2(a2 – x2) =1/2 x 200 x 22(102 – 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 22 x 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
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