Question

A 16 pound weight attached to a spring exhibits simple harmonic motion. Determine the equation of motion if the spring constant is 8 lb/ft and if the weight is released 6 inches below the equilibrium position with a downward velocity of 1 ft/s.

Answer 1

Answer:

T = 2.82π s

x = Acos(0.71t)

Explanation:

This problem can be solved by using the expressions

$T = 2\pi \sqrt{\frac{m}{k} }$    ( 1 )

$x=Acos(\omega t) = Acos(\frac{2\pi }{T}t )$   ( 2 )

where T is the period of oscillation of the system, m is the mass of the object attached to the spring, k is the spring constant and x is the position of the object.

By replacing in the expression (1):

$T=2\pi \sqrt{\frac{16 lb}{8lb/ft}} = 2.82\pi s$

Taking 6 inches as the amplitude of the motion, we have

$x=6cos(\frac{2\pi }{2.82\pi }t ) = 6cos(0.71t)$

I hope this is useful for you

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