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Mechanical Properties Of Fluids

Question
CBSEENPH11019185

A particle executes simple harmonic motion with amplitude 0.6m. Its velocity at mean position is 12m/s. Find the frequency of oscillation and acceleration at extreme position.

Solution

Given that,
Amplitude of the motion, r= 0.6cm
Velocity at mean position, v= 12m/s
We space know comma space straight nu subscript straight omicron equals rω
Therefore, '

space space space space space space space space space space space space space straight omega equals straight v subscript straight omicron over straight r equals fraction numerator 12 over denominator 0.6 end fraction equals 20 rad divided by straight s
and
space space space space space space space space space space space space space straight f equals fraction numerator straight omega over denominator 2 straight omega end fraction equals fraction numerator 20 over denominator 2 straight pi end fraction equals 3.183 Hz
The acceleration of particle at extreme position is, 
straight alpha subscript straight omicron equals r omega squared
space space space space equals 0.6 left parenthesis 20 right parenthesis squared
space space space space equals 240 c m divided by straight s squared