At what displacement is the ke of a particle performing shm of amplitude 10cm, three times its pe?

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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:

    At what displacement is the ke of a particle performing shm of amplitude 10cm, three times its pe?

    At what displacement is the ke of a particle performing shm of amplitude 10cm, three times its pe?

    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

    At what displacement is the ke of a particle performing shm of amplitude 10cm, three times its pe?

    ∴   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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