1.

The normal length of a steel spring is 8cm. Keeping one end of the spring fixed at a point, if a weight is attached to the other end, its length becomes 14 cm. The weight is pulled down slightly and then released. Find the time period of oscillation of the spring?

Answer»

Solution :Increase in LENGTH of the spring for the mass m is,
l = 14 - 8 = 6 cm.
So, FORCE constant k (force required for a unit increase in length) = `(mg)/6"DYN"*cm^(-1)`.
Time PERIOD, `T=2pisqrt(m/k)=2pisqrt(m/((mg)/6))`
= `2pisqrt(6/g)=2xx3.14xxsqrt(6/980)=0.49s`.


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