Solving sec^-1 in Horizontal Displacement Equation

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In summary: Originally posted by Hurkyl Sec = secondsHz = cycles per seconds^{-1} = angular velocity (radians/second)
  • #1
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I've got a problem where I'm given some variables that I don't understand the value of. The formula is for horizontal displacement of damped oscillating objects. The values I'm not understanding are:

(beta) = 0.1 sec^-1
(omega) = .05 sec ^-1

What does sec represent? seconds? secant?

I thought maybe it was .1 seconds raised to the negative 1 power, but the graph I get doesn't match the answer. I'm trying to put the formula into an excel spreadsheet. The full equation is:

x=x(naught)e^(-beta*time)*[cos(omega*time)+(beta/omega)sin(omega*time)]

I would appreciate any help on this.

Thx
 
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  • #2
In general

[tex]x^{-1} = 1/x[/tex]

In particular,

[tex]\mathrm{sec}^{-1} = 1 / \mathrm{sec} = \mathrm{Hz}[/tex]

(yes, sec = seconds)
 
  • #3
As Hurkyl said, it is seconds

Ask yourself this, if it was secant...then wouldn't there have to be a following value?? or are you going to take a secant of nothing?
 
  • #4
Thanks, Hurkyl!
 
  • #5
Just a technical point to avoid confusion: Both sec^-1 and Hz are used to denote frequency. However, by convention, Hz stands for cycles per second and sec^-1 is the angular rate (radians per second): they differ by a factor of [itex]2\pi[/itex]. In particular,

[tex]2\pi\mbox{ sec}^{-1}=1\mbox{ Hz}[/tex]

Edit: corrected as per NateTG post.
 
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  • #6
Originally posted by krab
Just a technical point to avoid confusion: Both sec^-1 and Hz are used to denote frequency. However, by convention, Hz stands for cycles per second and sec^-1 is the angular rate (radians per second): they differ by a factor of [itex]2\pi[/itex]. In particular,

[tex]1\mbox{ sec}^{-1}=2\pi\mbox{ Hz}[/tex]

Don't you mean [tex]2\pi s^{-1}=1 \mbox{Hz}[/tex]? 1 Hertz is a cycle per second which is [tex]2\pi[/tex](radians) per second.

I usually think of [tex]s^{-1}[/tex] as being a unit of angular velocity, and [tex]Hz[/tex] as a unit of frequency.

Of course, since [tex]2\pi[/tex] is unitless, there can be multiple definitions of [tex]s^{-1}[/tex].

P.S. [tex]s^{-1}[/tex] is often read as 'per second.'
 
  • #7
Originally posted by NateTG
Don't you mean [tex]2\pi s^{-1}=1 \mbox{Hz}[/tex]?
Sorry. My bad. I'll fix it (like revising the congressional record). I'll attribute it to you so it's clear why you corrected it.
 

1. What does "sec^-1" mean in the horizontal displacement equation?

The notation "sec^-1" in the horizontal displacement equation represents the inverse of the secant function. This function is used to calculate the angle of a right triangle given the length of its adjacent side and hypotenuse.

2. How do you solve for "sec^-1" in the horizontal displacement equation?

To solve for "sec^-1" in the horizontal displacement equation, you can use a scientific calculator or a trigonometric table. Simply input the value of the adjacent side and hypotenuse and use the inverse secant function to calculate the angle.

3. Why is "sec^-1" important in the horizontal displacement equation?

The inclusion of "sec^-1" in the horizontal displacement equation is important because it allows us to calculate the angle of a right triangle without knowing the length of the opposite side. This is especially useful in real-world applications, such as navigation and engineering.

4. Can you solve for "sec^-1" using the Pythagorean theorem?

No, the Pythagorean theorem only applies to right triangles and cannot be used to calculate the angle in the horizontal displacement equation. The inverse secant function must be used to find the angle in this equation.

5. Are there any common mistakes when solving for "sec^-1" in the horizontal displacement equation?

Yes, one common mistake is forgetting to use the inverse function when solving for "sec^-1". It is also important to make sure the calculator is in the correct mode, either degrees or radians, when using the inverse secant function.

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