Simple harmonic motion [ I have an exam tomorrow morning]

[Telemachus]

Simple harmonic motion [URGENT I have an exam tomorrow morning]

Homework Statement

Hi there. I don't know how to solve this problem about simple harmonic motion.
It says: A particle of 100 grams is animated with a simple harmonic motion, which starts 20cm from its equilibrium position, where it gets a speed of 200cm/s.
a) Calculate the period and the frequency of the oscillation.

Homework Equations

x(t)=A\sin(wt)+B\cos(wt)

The Attempt at a Solution

Well, I have a problem with finding the angular frequency. I need the "K", the coefficient of the spring (its a way of thinking it as a mass attached to a spring). And I don't know how to find it.

Any ideas?

Bye there!

 

[Twoism]

It's possible to find the K value.
The total mechanical energy of the system has a constant value equal to the potential and kinetic energy of the system.
K + U = E
In SHM, Energy is also equal to .5kA^2
That would mean...
K+U=.5kA^2
Is there a way to make the kinetic energy or potential energy equal to zero so all you have is
K=.5kA^2
or
U=.5kA^2

What is U and K equal to? Can you find them with the given information?

 

[Telemachus]

I wanted to do it using the solution for the differential equation. We don't make energetic considerations yet.

Thank you for helping anyway.

 

[berkeman]

Homework Statement

Hi there. I don't know how to solve this problem about simple harmonic motion.
It says: A particle of 100 grams is animated with a simple harmonic motion, which starts 20cm from its equilibrium position, where it gets a speed of 200cm/s.
a) Calculate the period and the frequency of the oscillation.

Homework Equations

x(t)=A\sin(wt)+B\cos(wt)

The Attempt at a Solution

Well, I have a problem with finding the angular frequency. I need the "K", the coefficient of the spring (its a way of thinking it as a mass attached to a spring). And I don't know how to find it.

Any ideas?

Bye there!

They give you both the amplitude and the speed at zero displacement... How do you get the speed from the equation you wrote?

 

[Liquidxlax]

whats the derivative of position as a function of time?

 

[Telemachus]

Taking the derivative.
If we considere x_0=x_e, this is, the initial point at the point of equilibrium, then we get:

x(t)=\frac{200cm}{w s}\cos(wt)

What I really need to find is the angular frequency, the omega (w in the equation).

w=\sqrt {\frac{k}{m}}

x(t)=A\sin(wt)+B\cos(wt)

Here A=x_0 and B=\frac{\dot x_0}{w}

 

[berkeman]

Taking the derivative.
If we considere x_0=x_e, this is, the initial point at the point of equilibrium, then we get:

x(t)=\frac{200cm}{w s}\cos(wt)

What I really need to find is the angular frequency, the omega (w in the equation).

w=\sqrt {\frac{k}{m}}

x(t)=A\sin(wt)+B\cos(wt)
Here A=x_0 and b=\frac{\dot x_0}{w}

I don't understand what you just wrote. You had it mostly right in the first post, but just simplify it to x(t) = B cos(wt).

You're given B, and you're given x'(t) when x=0. Do the differentiation and ...

 

[Telemachus]

I've just replaced B for B=\frac{\dot x_0}{w} and x_0=0

If I differentiate then:

x(t)=\frac{200cm}{w s}\cos(wt)

x'(t)=-\frac{200cm}{s}\sin(wt)

 

[berkeman]

I've just replaced B for B=\frac{\dot x_0}{w} and x_0=0

If I differentiate then:

x(t)=\frac{200cm}{w s}\cos(wt)

x'(t)=-\frac{200cm}{s}\sin(wt)

I could be wrong, but the term B would seem to be simpler than what you have written above

which starts 20cm from its equilibrium position

And what is the "s" term in the denominator? Do the units of your B come out to cm?

 

[Telemachus]

The s is for seconds, and cm centimeters. I've changed the reference point to the equilibrium point. I think its simpler that way cause it oscillates around that point.

 

[berkeman]

The s is for seconds, and cm centimeters. I've changed the reference point to the equilibrium point. I think its simpler that way cause it oscillates around that point.

I hope I'm not being dense, but I'm still not tracking. What does it mean to write "seconds" in the denominator? The units of omega are not seconds. How do the units of your B term come out to be just length (cm)?

I need to bail for a while, but will try to check back later. Maybe try it my way (post #7) to see if you get to the answer sooner.

 

[Telemachus]

Its because it tells me that the speed is 200cm/s at the equilibrium point. I could be wrong, maybe you're right. I don't find the way to get the omega.

 

[berkeman]

Its because it tells me that the speed is 200cm/s at the equilibrium point. I could be wrong, maybe you're right. I don't find the way to get the omega.

In the equation x(t) = B cos(wt), the units of x are cm, so the units of B have to also be distance, right? B is the maximum half-amplitude of the sinusoidal SHM, right?...

 

[Telemachus]

Right.

I've found it :P

w=\frac{10}{s}

 

[berkeman]

Cool. Have a good exam tomorrow.