Calcultating circular frequency and phase on a spring

[Acookook]

Am I trying to solve a simple equation, but get stuck at the basic. I have have a spring with a cube attached to it (1kg). When I pull it with 5N, the spring extends to 50mm =amplitude, (Hooks law) the koefitient is 120. The determined absortion coeffitient is 0,2^-1s. I tried to get the circular frequency, phase in the moment of releasing the cube, but i am not even getting the right results 9,99rad/s, and the phase is -1.15°. The formula ω=sqrt(k/m) doesn't give the right answer even when calc is in rad mode. If we'd like to know the force,what step would be taken?

Formulas
1. F=k/s -> F= force, k = coeffitient, s= distance
2. x(t)=A*cos(ωt+ρ)

3. The Attempt at a Solution
I tried to use the formula for the circular frequency (rad/s), but couldn't get past this one as I don't get the right answer,even when calc is in RAD mode. I am using this the formula ω=sqrt(k/m).

 

[kuruman]

Hi Acookook and welcome to PF.

When I pull it with 5N, the spring extends to 50mm =amplitude, (Hooks law) the koefitient is 120.

How did you figure 120? What are the units?

The determined absortion coeffitient is 0,2^-1s.

What is this quantity? Please explain.

 

[Acookook]

The cube itself is 1kg =10N, we tow it with 5N (so 15N in total). I misscalculated 120, but in general I can use hooks law to calculate the k=F/s (50milimeters).
The absortion coffetient determines how much sound is absorbed. If the abs. koef = 1, then no sound is reflected.

 

[kuruman]

The cube itself is 1kg =10N, we tow it with 5N (so 15N in total). I misscalculated 120, but in general I can use hooks law to calculate the k=F/s (50milimeters).

The weight of the cube is irrelevant. Imagine the mass on a frictionless table attached to one end of the spring. You pull at the other end of the spring with a constant force of 5 N and as you pull, you see the mass accelerate with the spring extended by 50 mm. Please answer the next questions, (a) What is the net force acting on the mass? (b) What is the force exerted by the spring on the mass? (c) What is the spring constant k according to Hooke's Law?

The absortion coffetient determines how much sound is absorbed. If the abs. koef = 1, then no sound is reflected.

What kind of sound is this and why do you worry about it? What is its connection to the problem? Are you confusing the frequency of oscillations $\omega$ with some kind of sound frequency?

 

[Acookook]

On the cube the force is acting is 5N and is extended by 50mm So the constant according to hook's law is F=kx => 5N/0,05m =100. The time absortion coefitient of the swing is 0.2 s ^-1. It also bothers me too, because In my school book the is no real refence. For the circular freq I have ω=sqrt(k/m)=(2π)/T.

 

[kuruman]

The time absortion coefitient of the swing is 0.2 s ^-1.

What swing? Is this spring swinging like a pendulum? I thought you were looking for the angular frequency of spring oscillations ω. If k = 100 N/m and m = 1 kg, what is ω = sqrt(k/m)? You don't need a calculator for this.

 

[Acookook]

I will ask my teacher what he meant with the 0.2s^-1 and post a clarified answer tommorow.

 

[kuruman]

I will ask my teacher what he meant with the 0.2s^-1 and post a clarified answer tommorow.

OK.

 

[Acookook]

Hi! Sorry, for the late reply, but I had other thing to do. I asked my teacher for the explanation, but he told me that I have everything to solve the problem. I talked to my classmates and they've we have written these formulas posted above, related it to the problem with the spring (drawing a sketch of an object connected to the spring), but never performed any calculations or something else. It likely won't be in the test. I'll try to look for some (un)solved problems regarding this problem and post it there.
We're taking the acoustic part of physics, so far we've covered calculation examples for a (guitar) string, plane waves, breaking of the waves and this stupid spring. I'll post an example ASAP.