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A spring block system is on a horizontal plane. If spring constant is 2N/m & mass of block is 1kg. By what amount should the block be extended from equilibrium position such that maximum velocity at mean position is 2m/s?(a) 2√2 m(b) 3√2 m(c) 4√2 m(d) 2 mI had been asked this question at a job interview.Question is from Oscillations topic in portion Oscillations of Physics – Class 11

Answer»

Right choice is (a) 2√2 m

The best EXPLANATION: Using, W = √ (k/m), we GET: w = √ (1/2)s^-1.Maximum VELOCITY = Aw.2 = A/√2∴A = 2√2 m.



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