Explanation:
It is given that,
Frequency of vibration, f = 215 Hz
Amplitude, A = 0.832 mm
(a) Let T is the period of this motion. It is given by the following relation as :



(b) Speed of sound in air, v = 343 m/s
It can be given by :




Hence, this is the required solution.
Yes. It's (speed squared)/(radius of the circle).
I think you would use F = ma
F = 65*10
F = 650N
(The 10m/s is from acceleration due to gravity)
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