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16 August, 00:40

At t=0, an 830-g mass at rest on the end of a horizontal spring (k = 169 N/m) is struck by a hammer which gives it an initial speed of 2.32 m/s.

Determine (a) the period and frequency of the motion, (b) the amplitude, (c) the maximum acceleration, (d) the total energy, and (e) the kinetic energy when

x = 0.40 A, where A is the amplitude.

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  1. 16 August, 01:07
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    Answer:a) T=0.44seconds and f=2.29Hz

    b) A=0.1626m

    c) amax=33.11m/s^2

    d) E=2.2337J

    e) KE=1.8732J

    Explanation:

    Period of the motion:

    T=2π√m/k

    T=period, m=mass of the body, k=spring constant

    Converting mass from gram to kilogram:

    M=830g/1000=0.83kg

    T=2π√0.83/169=0.44038seconds

    T=0.44secs (approximately)

    Frequency of the motion:

    F=1/T

    Where T=0.44038seconds

    F=1/0.44038s=2.27Hz

    b) maximum speed:

    Vmax=A√k/m

    Where A=amplitude, m=mass of the body, k=the spring constant

    Making A the subject of the formula:

    A=Vmax√m/k

    Where, speed (V) = 2.32m/s as Vmax and 0.83kg for m and 169N/m for k.

    A = (2.32m/s) √ (0.83kg) / (169N/m) = 0.1626m

    c) Maximum acceleration:

    amax = A (k/m)

    Therefore, A=

    (0.1626m) (169N/m) / 0.83kg)

    A=33.11m/s^2

    d) Total Energy:

    E=1/2MVmax

    Where, M=0.83kg and Vmax=2.32m/s

    E=1/2 * (0.83kg) * (2.32m/s) ^2

    E=2.2337J

    e) From the law of conservation of mass:

    1/2kA^2=1/2Kx^2+KE

    Where x=0.04A

    KE=1/2kA^2-1/2kx^2

    KE=1/2kA^2 (1-0.4^2)

    KE=Etotal*0.84

    Where, Etotal=2.23J

    KE=2.23*0.84=1.8732J
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