Want in detail A spring fixed at the upper end supports a weight of 980 gm at its lower end.cm under a load of 10 gm and the resistance (in gm wt.) to the motion ofThe spring stretchesthe weight is numerically equal to of the speed of the weight in cm/sec. Theweight is pulled down cm. below its equilibrium position and then released.Find the expression for the distance of weight from its equilibrium position attime t during its first upward motion.Also find the time it takes the damping factor to drop to of its initial value.10

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A spring fixed at the upper end supports a weight of 980 gm at its lower end. The spring stretches cm under a load of 10 gm and the resistance (in gm wt.) to the motion of 2 the weight is numerically equal to of the speed of the weight in cm/sec. The weight is pulled down ! Find the expression for the distance of weight from its equilibrium position at time t during its first upward motion. 10 cm. below its equilibrium position and then released. to Also find the time it takes the damping factor to drop to of its initial value. 1 10

A spring fixed at the upper end supports a weight of 980 gm at its lower end. The spring stretches cm under a load of 10 gm and the resistance (in gm wt.) to the motion of 2 the weight is numerically equal to of the speed of the weight in cm/sec. The weight is pulled down ! Find the expression for the distance of weight from its equilibrium position at time t during its first upward motion. 10 cm. below its equilibrium position and then released. to Also find the time it takes the damping factor to drop to of its initial value. 1 10

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter7: Dry Friction

Section: Chapter Questions

Problem 7.78P: The figure shows a steel bar being processed by a rolling mill. Given that P=80kN and r =0.016,...

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