What is the Unit of ω=sqrt(g/x0) and How is it Converted to Radians?

In summary, the conversation is about a question related to physics and a solution provided in a book. The question asks for an estimate of the oscillation frequency of a heavy object placed on a rubber pad as a shock absorber. The book gives a solution of sqrt(980) rad/s, and the question is about how the units of meters, seconds, and centimeters are converted to radians for the value of ω=sqrt(g/x0).
  • #1
Barrowlands
4
0
So, I'm trying to brush up on my undergrad physics, and I'm sure this is a bone-headed question, so please bear with me.

Homework Statement


A heavy object, when placed on a rubber pad that is to be used as a shock absorber, compresses the pad by 1cm. If the object is given a vertical tap, it will oscillate. Ignoring the damping, estimate the oscillation frequency. [The book I'm using actually gives the solution]

Homework Equations


x(t)=A*sin(sqrt(k/m)t+[itex]\phi[/itex])
ω=sqrt(k/m)
F=k|l-l0|

The Attempt at a Solution


We'll call x0 the equilibrium displacement, x0=1cm
k=spring constant of rubber
so
k(l-l0)=k*x0=mg (equilibrium)
gives us
k=(mg)/(x0)
then
ω=sqrt(k/m)
which eventually solves to
ω=sqrt(g/x0)

The book gives an answer of sqrt(980) rad/s. My question is given the units from ω=sqrt(g/x0) (meters, seconds, centimeters), how do they arrive at radians?

Edit: I just realized I put this in the wrong sub-forum, my apologies.
 
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  • #2
The unit of sqrt (g/x0) is s^-1. A "radian" is not really a unit, it is merely an indication that the corresponding dimensionless value is used as an angular measure.
 

FAQ: What is the Unit of ω=sqrt(g/x0) and How is it Converted to Radians?

1. What is an ideal spring?

An ideal spring is a theoretical model of a spring that is massless, frictionless, and has a linear relationship between the force applied and the displacement of the spring.

2. What are the units of an ideal spring?

The units of an ideal spring depend on the equation being used. For Hooke's Law, the units are typically Newtons per meter (N/m). For the spring constant, the units are typically Newtons per meter per meter (N/m2).

3. How is an ideal spring different from a real spring?

An ideal spring is a simplified model that does not account for factors such as the mass of the spring, friction, and non-linear relationships between force and displacement. In reality, these factors can affect the behavior of a spring and make it deviate from the ideal model.

4. How do you calculate the force of an ideal spring?

The force of an ideal spring can be calculated using Hooke's Law, which states that the force (F) is equal to the spring constant (k) multiplied by the displacement (x) from its equilibrium position. Mathematically, this can be represented as F = kx.

5. What are some real-life applications of ideal springs?

Ideal springs are used in various mechanical systems, such as in car suspensions, door hinges, and shock absorbers. They are also commonly used in physics experiments and simulations to model the behavior of springs in ideal conditions.

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