Determine the frequency of the oscillatory movement if the oscillation period is: 0.02 s, 0.4 s, 5 s, 50 s, 1 h
| August 29, 2021 | education
T1 = 0.02 s.
T2 = 0.4 s.
T3 = 5 s.
T4 = 50 s.
T5 = 1 h = 3600 s.
v1 -?
v2 -?
v3 -?
v4 -?
v5 -?
The oscillation period T is the time of one complete oscillation. The oscillation period T is determined by the formula: T = t / N, where t is the time during which the pendulum makes N oscillations.
The frequency of oscillations v is the number of oscillations per unit of time. The frequency is determined by the formula: v = N / t.
It can be seen from the formulas that the frequency v is the reciprocal of the period T: v = 1 / T.
v1 = 1 / 0.02 s = 50 Hz.
v2 = 1 / 0.04 s = 25 Hz.
v3 = 1/5 s = 0.2 Hz.
v4 = 1/50 s = 0.02 Hz.
v5 = 1/3600 s = 0.00028 s.
Answer: v1 = 50 Hz, v2 = 25 Hz, v3 = 0.2 Hz, v4 = 0.02 Hz, v5 = 0.00028 s.
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