Question

In: Advanced Math

will rate :-) A mass weighing 8 pounds stretches a spring 2 feet. At t=0 the...

will rate :-)

A mass weighing 8 pounds stretches a spring 2 feet. At t=0 the mass is released from a point 2 feet above the equilibrium position with a downward velocity of 4 (ft/s), determine the motion of the mass.

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Expert Solution

Hello!!

According to Hooke's law, if we stretch a spring by displacing it's free end by , then, the restoring force is set up in opposite direction to overcome stretching, which is opposite direction to the displacement.

If we don't concern the direction then, magnitude of restoring force, . So, removing proportionality yields .

Now, see the figure below:

When mass is not placed on the spring and spring is massless it will be in equilibrium. When a mass , m is placed at free end of spring, the spring is stretched by say x units. In this case, we have two forces one is spring's restoring force, and it is in upward direction while the other is weight, downwards. Since, block is at equilibrium, so, the block is stationary, i.e., both forces cancel out each other.

Thus, .

Here, , and , which gives

.

Further, if the block is released from 2 ft above the equilibrium then, the motion will be as follows (Observe direction of velocities at extreme ends).

This motion will repeat itself and is called simple harmonic motion.

Note: If , then, the net force is , considering direction of both force and displacement. Also, according to Newton's second law of motion, we have . In this case, only restoring force is external force due to displacement from equilibrium.

So, .

Using the value of and , we get

  

The sol. of this equat. is of the form , where . Here, assume the displacement above the equilibrium is positive while the displacement below it is negative. [Note : x = 0 at equilibrium ]. Now, we need to determine values of .

At , substituting the values gives [ EQUATION 1 ]

As, velocity,

and we have at , so, substituting the value gives

[ EQUATION 2 ]

Also, acceleration, , here, direct comparison with differential equation gives .

Substituting in equation 2 gives, [ EQUATION 3 ]

Now, squaring and adding equation 1 and equation 3, we get

.

Substitute the value of in equation 1 and 3 to get,

and .

,

Therefore, we finally get . This equation gives position of the block at any time, t.

I have kept everything as descriptive as possible

But still if you get any doubt leave comment

Please give like if my work helps.


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