1.

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.

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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