Physics, asked by Hopez, 9 months ago

A simple pendulum of weight W and length l is released from rest at A as shown in figure. It strikes a spring of stiffness k at B. Calculate the deflection of the spring.

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Answered by Natsukαshii
0

Imagine a car parked against a wall. If a bulldozer pushes the car into the wall, the car will not move but it will noticeably change shape. A change in shape due to the application of a force is a deformation. Even very small forces are known to cause some deformation. For small deformations, two important things can happen. First, unlike the car and bulldozer example, the object returns to its original shape when the force is removed. Second, the size of the deformation is proportional to the force. This second property is known as Hooke’s law. In equation form, Hooke’s law is

F=−kx

where x is the amount of deformation (the change in length, for example) produced by the restoring force F , and k is a constant that depends on the shape and composition of the object. The restoring force is the force that brings the object back to its equilibrium position; the minus sign is there because the restoring force acts in the direction opposite to the displacement. Note that the restoring force is proportional to the deformation x. The deformation can also be thought of as a displacement from equilibrium. It is a change in position due to a force. In the absence of force, the object would rest at its equilibrium position. The force constant k is related to the stiffness of a system. The larger the force constant, the stiffer the system. A stiffer system is more difficult to deform and requires a greater restoring force. The units of k are newtons per meter (N/m). One of the most common uses of Hooke’s law is solving problems involving springs and pendulums, which we will cover at the end of this section.

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