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16 August, 03:04

Frequency and amplitude of a particle in simple harmonic motion A particle moves in simple harmonic motion. Knowing that the maximum velocity is 200 mm/s and the maximum acceleration is 13 m/s, determine the amplitude and frequency of the motion. Amplitude of the motion xm 1mm. Frequency of the motion勿 Hz.

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  1. 16 August, 03:11
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    amplitude, a = 3.076 mm

    frequency, f = 10.35 Hz

    Explanation:

    Vmax = 200 mm / s = 0.2 m/s

    Amax = 13 m/s

    Let f be the frequency and a be the amplitude.

    Use the formula of maximum velocity.

    Vmax = ω a

    0.2 = ω a ... (1)

    Use the formula of maximum acceleration.

    Amax = ω^2 x a

    13 = ω^2 a ... (2)

    Divide equation (2) by (1)

    ω = 13 / 0.2 = 65 rad/s

    Put in equation (1)

    0.2 = 65 x a

    a = 3.076 x 10^-3 m

    a = 3.076 mm

    Let f be the frequency

    ω = 2 π f

    f = 65 / (2 x 3.14) = 10.35 Hz
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