Simple Harmonic Motion

Printable View

August 25th 2006, 11:01 AM
spiritualfields

Simple Harmonic Motion

There's probably a simple explanation that I'm overlooking, but I can't see it. This is a problem dealing with interpretation of something in my text book, so I'll copy it verbatim, then ask my question. First, simple harmonic motion is described as:

Quote:

Simple harmonic motion is a special kind of vibrational motion in which the acceleration a of the object is directly proportional to the negative of its displacement d from its rest position. That is,
a = -kd.

With that in mind, the text goes on to elaborate further, calling attention to a figure that has an object suspended from a spring. The object has a rest position (point B), a maximum up displacement position (point A), and a maximum down displacement position (point C). The text says:

Quote:

For example, when the mass hanging from the spring in Figure 43 is pulled down from its rest position B to the point C, the force of the spring tries to restore the mass to its rest position. Assuming that there is no frictional force to retard the motion, the amplitude will remain constant. The force increases in direct proportion to the distance that the mass is pulled from its rest position. Since the force increases directly, the acceleration of the mass of the object must do likewise, because (by Newton's Second Law of Motion) force is directly proportional to acceleration. As a result, the acceleration of the object varies directly with its displacement, and the motion is an example of simple harmonic motion.

The problem is that there is also a figure that shows the sinosoidal shape of the displacement (from rest) with respect to time. The sinosoidal waveform seems to demonstrate the exact opposite of what the text is asserting! Specifically, the object is exhibing the MOST acceleration when the displacement is least (near rest). I corrolate the slope of the sinosoidal waveform with the acceleration of the object, and the slope changes the most when the object is moving through its rest position. The acceleration is LEAST when the object is at maximum displacement (corresponding to the maximum and minimum parts of the sinosoidal, when the slope slows to near zero).

There seems to be a disconnect between what the sinosoidal waveform is showing and what the description is saying, in regards to acceleration being directly proportional to displacement. It seems that acceleration of the object is GREATEST when it is passing through rest. I mean I can't see any other way to interpret the sinosoidal waveform which is graphing the object's movement!

Hopefully, I've provided enough information to give an idea of what is confusing me. If not, I'll have to figure out how to post a diagram.
August 25th 2006, 11:33 AM
CaptainBlack

Quote:

Originally Posted by spiritualfields

There's probably a simple explanation that I'm overlooking, but I can't see it. This is a problem dealing with interpretation of something in my text book, so I'll copy it verbatim, then ask my question. First, simple harmonic motion is described as:

With that in mind, the text goes on to elaborate further, calling attention to a figure that has an object suspended from a spring. The object has a rest position (point B), a maximum up displacement position (point A), and a maximum down displacement position (point C). The text says:

The problem is that there is also a figure that shows the sinosoidal shape of the displacement (from rest) with respect to time. The sinosoidal waveform seems to demonstrate the exact opposite of what the text is asserting! Specifically, the object is exhibing the MOST acceleration when the displacement is least (near rest). I corrolate the slope of the sinosoidal waveform with the acceleration of the object, and the slope changes the most when the object is moving through its rest position. The acceleration is LEAST when the object is at maximum displacement (corresponding to the maximum and minimum parts of the sinosoidal, when the slope slows to near zero).

The slope of the displacement is the velocity, which is a maximum when
the dispalecement is zero.
August 25th 2006, 12:28 PM
spiritualfields

Quote:

The slope of the displacement is the velocity, which is a maximum when
the dispalecement is zero.

Yes, but isn't the rate of change of velocity directly proportional to acceleration? In other words, if acceleration is zero, then the velocity is constant. If acceleration is maximum, then the velocity is changing rapidly. In order for the rate of change of the velocity of the object to be greatest when it is near its rest position, wouldn't that also mean that the object would also have to be undergoing its greatest acceleration?
August 25th 2006, 12:36 PM
CaptainBlack

Quote:

Originally Posted by spiritualfields

Yes, but isn't the rate of change of velocity directly proportional to acceleration? In other words, if acceleration is zero, then the velocity is constant. If acceleration is maximum, then the velocity is changing rapidly. In order for the rate of change of the velocity of the object to be greatest when it is near its rest position, wouldn't that also mean that the object would also have to be undergoing its greatest acceleration?

The slope of displacement, when the displacement is zero, is the velocity
and is a maximum, but the rate of change of slope, which is the accelleration
is zero at this point. Which is as it should be as the accelleration is zero.

RonL
August 25th 2006, 01:23 PM
spiritualfields

Quote:

The slope of displacement, when the displacement is zero, is the velocity
and is a maximum

That I understand. Now this:

Quote:

but the rate of change of slope, which is the accelleration
is zero at this point. Which is as it should be as the accelleration is zero.

In other words, because near rest the slope (velocity) is effectively linear and therfore not changing (and thus not accelerating)? I can see this now, but it certainly was not easy to pick out from the text book description. One of the disadvantages, I suppose, that comes with teaching yourself. Thanks.