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A light vertical spring is stretched by 0.2 cm when a weight of 10 g is attached to its free end. The weight is further down by 1.0 cm and released. Compute the frequency and maximum velocity of load. |
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Answer» Solution :(i) Force CONSTANT of the spring `k=("Restoring Force")/("Increase in length")=(mg)/("Increase in length") ( :. F=mg)` `=(10^(-2)xx9.8)/(2.0xx10^(-3))=49Nm^(-1)` Frequency `v=1/(2pi)SQRT(k/m)=1/(2pi)sqrt(49/(10^(-2)))=1/(2pi)xx70` (II) The amplitude of motion = The distance through which the WEIGHT is further pulled down =1.0cm i.e. `A=1.0cmimplies` Maximum velocity `=AW=10^(-2)mxx1/70"rad"s^(-1)=1/7000ms^(-1)` |
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