Question

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.

Answer 1

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

Answer 2

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.