A block is suspended by a spring and the elongation produced in the spring is found to be x0. Then the block is further pulled downward by a distance x0 and released. Then acceleration of the block in its lower most position is
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a
g
b
g/2
c
2g
d
zero
answer is A.
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Detailed Solution
After release, the block will execute vertical simple harmonic motion about is equilibrium position with amplitude (2x0−x0) i.e. x0 and angular frequency ωHere ω=km and x0=mgk∴ Acceleration at the lower extreme position =ω2x0=(Km) ×mgk=g