Single Degree of Freedom Vibration
16 Pages1187 Words243 Views
Added on 2019-10-31
Single Degree of Freedom Vibration
Added on 2019-10-31
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Vibration1 | PageSingle Degree of Freedom Vibration
VibrationContentsSolution 1.........................................................................................................................................3Solution 2.........................................................................................................................................4Solution 3.........................................................................................................................................5Solution 4.........................................................................................................................................6Solution 5.........................................................................................................................................8Solution 6.........................................................................................................................................9Solution 7.......................................................................................................................................11Solution 8.......................................................................................................................................132 | Page
VibrationSolution 1We can assume that, If the steel ball is falling from height h, It has potential energy mgh, which is concentrated in to kinetic energy, when the ball will touch the tabletop it velicity will bemgh=12mv2orv=√2ghIf the horizontal table top is fixed and the coefficient of restitution is zero then the ball will bounce back to the same height. But here is spring with stiffness constant K plays important role.The energy is loosed due to expansion and contraction of the spring. When the ball collide with table top,If we consider,e=v2−v2u2−u1 (This is also known as coefficient of restitution)u1⟶Velocity of the ballu2⟶ Velocity at table top = 0v1⟶Velocity of the ball after collision.v2⟶Velocity of the table top after collisione=v2−v2u2The velocity of the ball after collision ev where v is the velocity of the steel ball after collision.0=(eυ)2−2gh(We know that v2 = u2 – 2gx)h=e2ν22g=e22gh2g=e2hTherefore,Total distance covered before stopd=h+2e2h+2e4h........∝3 | Page
Vibration¿1+e21−e2hThe velocity of the ball decreases due to the loss of energy because of the sping.Solution 2As given in question,Ft = change in momentumFt = mVf – mVi = 400 x 0.003 = 4.7 VfVf = 0.255 m /secThe total work done will be given as Work done = Δ K.E. (Kinetic Energy)⟹2000(x)2=4.7x(0.255)2X (Deflection) = 0.0123 m = 12.377 mmWe know that,F = ma⟹ 400 = 4.7 amax⟹ amax = 85.106 m / sec2Ans4 | Page
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