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The particle executing SHM in a straight line has velocities 8 m/s p 7 m/s and 4 m/s at three points having distance 1 m from each other. what will be maximum velocity of particle ? |
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Answer» imum velocity of the particle will be √65 m/sExplanation : The maximum velocity of a particle executing SHM is GIVEN by,Vmax = AwThe velocity at any point X is given by,V² = w²(A²-x²)where w = angular velocity,A = amplitudex = distance from central positionlet the three points A,B,C are at a distance of d, d+1, d+2 meters from the central position,hence,V₁² = w²(A²-d²) = 64.............eq1V₂² = w²(A²-(d+1)²) = 49.............eq2V₃² = w²(A²-(d+2)²) = 16............eqn3subtracting eqn2 from eqn1 we get15 = w²(2d+1).................eqn4subtracting eqn3 from eqn2 we get33 = w²(2d+3)...........eqn5dividing eqn5 and eqn4 we get(2d+3)/(2d+1) = 33/15=> 30D + 45 = 66d + 33=> d = 1/3putting the value of d in eqn4 we get,15 = w²(2/3 + 1 )=> w = 3putting the value of w and d in eqn1 we getw²(A²-d²) = 64=> 9(A²-1/9) = 64=> A² = 64/9 + 1/9 = 65/9=> A = √65/3Hence maximum velocity of the particle,Vmax = Aw = √65 |
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