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16 December, 20:30

The tip of a tuning fork goes through 440 complete vibrations in a time of 0.510s. Find the angular frequency and the period of the motion.

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Answers (2)
  1. 16 December, 21:23
    0
    44 complete vibrations in 0.510s.

    1 complete vibration would be in: 0.510/44 = 0.01159 s.

    Hence the Period, T = 0.01159 s

    Frequency, f = 1/T = 1/0.01159 ≈ 86.28 Hz

    Angular frequency, ω = 2πf = 2π*86.28 ≈ 542.11 rad/s
  2. 16 December, 22:52
    0
    Solve for the time it will take to complete a revolution. That is,

    0.510 s / 440 revolutions = 51/44000 s

    The frequency in Hertz is the reciprocal of this value thus the frequency is approximately equal to 862.75 Hz.

    Angular velocity expressed in radians/second

    (440 rev / 0.510 s) x (2π rad / 1 rev) = 5420.787 rad/s

    The period is the reciprocal of frequency which is approximately equal to 1.16x10^-3 s.
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