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A block of mass one kg is fastened to a spring with a spring constant `50Nm^(-1)`. The block is pulled to a distance `x=10cm` from its equilibrium position at `x=0` on a frictionless surface from rest at `t=0. ` Write the expression for its x(t) and v(t). |
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Answer» Here, `m=1kg,k-50Nm^(-1)` `a=10cm=0.10m` `omega=sqrt((k)/(m))=sqrt((50)/(1))=7.07rad//s` Since the motion starts from th emean position, so the displacement equation can be given as `x(t)=asinomegat=-.10sin7.07tm` and velocity, `v(t)=dx//dt=aomega cosomegat` `=0.10xx7.07 cos 7.07t ` `=0.707cos7.07tms^(-1)` |
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