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A particle executes simple harmonic motion with an amplitude of 4 cm At the mean position the velocity of the paricle is10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is(B) 5cm(C) 23)cm(D) 21V5)cm3 cm |
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Answer» As we know in SHM100=16 Omega,hence Omega=100\16(from given value) Now,Again ,5=100/16√A(square)-X(square) Hence X=3.37cm from mean position |
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