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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? |
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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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