To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion
From the definition we know that the frequency can be expressed as

Where,


Therefore the frequency would be given as


The frequency is directly proportional to the angular velocity therefore



Now the maximum speed from the simple harmonic movement is given by

Where
A = Amplitude
Then replacing,


Therefore the maximum speed of a point on the string is 3.59m/s
Radial acceleration is given by

where

then

Now

Using the relation


Putting into rpm
If there is a large enough amount of water vapor in the air and the tempature is hot enough cloud or storms can form.