a 4kg roller is attached to a spring of force constant 100N/m. it slides over a frictionless horizontal road. the roller is displaced from equilibrium position by 10cm and then released. its maximum speed will be?
2 answers
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A=0.1 m
ω=sqrt(k/m) =sqrt(100/4) =5 rad/s
v(max) =Aω =0.1•5=0.5 m/s
ω=sqrt(k/m) =sqrt(100/4) =5 rad/s
v(max) =Aω =0.1•5=0.5 m/s
1 answer