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Mechanical Properties Of Fluids

Question
CBSEENPH11019308

A spring of force constant 1200 N/m is mounted on a horizontal table as shown. A mass of 30 kg is attached to the free end of the spring, pulled side ways to a distance of 2-0 cm and released.

(a) What is the frequency of oscillation of the mass?

(b) What is the maximum acceleration of mass?

(c) What is the maximum speed of mass?

Solution

We have,
Spring constant, k=1200 N/m
Mass of the spring, m=30kg
Amplitude of vibration = 2.0 cm =0.02m
(a) We know frequency of oscillation is,
straight v equals fraction numerator 1 over denominator 2 straight pi end fraction square root of straight k over straight m end root equals fraction numerator 1 over denominator 2 straight pi end fraction square root of 1200 over 3 end root equals fraction numerator 20 over denominator 2 straight pi end fraction equals 10 over straight pi Hz
(b) We know maximum acceleration of mass is, space space space a equals straight omega squared straight A
therefore space straight a equals left parenthesis 2 πv right parenthesis squared space straight A equals open parentheses 2 straight pi 10 over straight pi close parentheses squared straight x 0.02 equals 8 space straight m divided by straight s squared 
(c) Maximum speed is,
V equals A omega equals A 2 pi v
space space space equals 0.02 cross times 2 pi cross times 10 over pi
space space space equals 0.4 space m divided by s