A 5.0-kg block, initially at rest on a level, frictionless surface, is attached

Question

A 5.0-kg block, initially at rest on a level, frictionless surface, is attached to a Hooke's Law spring with spring constant k= 80 N/m. The spring, which is also level, is rigidly attached to a wall on the other end as shown in the diagram above. Also assume that there is no friction in the spring or at the point of attachment. The block is then streched from its equilibrium position at x= 0.00 m to a distance of 50 cm to the right and released. Taking the initial time to be t = 0.00s at the release point, answer the following:

What is the amplitude of the resultant simple harmonic motion?

Explanation / Answer

initially spring is elongated from x=0 to x=50 cm in the x direction . energy stored in the block at that position(x= 50 cm) is only potential energy and kinetic energy is zero.so this position is one of the extreme position of the SHM. distance of this extreme position from the x=0 is called the amplitude of the SHM. so amplitude is 50 cm

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