A block is suspended by a spring and the elongation of the spring is found to be 10 cm. The block is now pulled downward by further 5 cm and released. Then frequency of oscillation of the block is (Take g = 10 m/s2)

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5π Hz

10π Hz

4π Hz

52π Hz

answer is A.

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Detailed Solution

If x0 be the static elongation produced in the spring, then x0=mg/K. Time period of oscillation, T=2πmK=2πx0g∴ Frequency f=1T=12πgx0=12π100.1Hz=5π Hz

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