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A mass of 2 kg is attached to the spring...

A mass of 2 kg is attached to the spring constant `50Nm^(-1)`. The block is pulled to a distance of 5cm from its equilibrium position at `x=0` on a horizontal frictionless surface from rest at `t=0`. Write the expression for its displacement at anytime t.

Text Solution

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Here m = 2kg, k = 50 N `m^(-1)`,
A = 5 cm ` 5 xx 10^(-2)m = 0.05m`
Since `omega = sqrt((k)/(m))`
`therefore omega = sqrt((50)/(2)) = 5 rads^(-1)`
Since ` x = A sin omegat`
`therefore x = 0.05 sin 5t`.
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