Building a Clock from a Spring: Exploring Potential Energy

[ikihi]

Homework Statement

You are in a spaceship far from any other objects, and you want to build a clock. You decide to build your clock out of a spring with a mass attached to it. You use a spring with spring constant k = 138 N/m, and you initially displace the mass a distance x=25.0 cm from equilibrium.

a) How much mass will you need so that your clock will measure seconds?
b) What is the max velocity of the attached mass?
c) Draw a plot of the spring potential energy (Usp) as a function of time for two periods of the clock. The plot must be neat and include numeric values for the max Usp. Make sure you also label the points when Usp is a min and a max, Give the numeric values for both the time and Usp.

Homework Equations

T= 2π ⋅ √m/k

V(max) = ± A ⋅ √k/m

The Attempt at a Solution

a) T²/ 4π²= m/k
m=(k * T^²)/(4 * π²)
m= 138 * 1 sec / ( 4 * π^2)
m= 3.50 kg (is this correct?)

b) V (max) = ± 0.25 m * √(138 N/m /3.496 kg) = ± 1.57 m/s (is this correct?)

c) I don't understand this part.

 

[Hiranya Pasan]

It should be periodic, since you know the solutions for harmonic oscillators (x=A sin(wt+e)) and the spring potential energy (U(x)=1/2kx^2) and just substitute ,
U(t)=1/2 k A^2 sin^2(wt+e) , now you can plot this function with numerical values.

 

[Cutter Ketch]

Can you write any equation for the potential energy? Seeing as you have determined the period and the amplitude can you write an equation for the displacement (x) as a function of time?

 

[Hiranya Pasan]

you can find e using the initial condition

 

[ikihi]

you can find e using the initial condition

what is e? The equation in my book just says x= A ⋅ cos (ω ⋅ t) and this goes to x = A ⋅ cos ( 2π/T ⋅ t)
And how do you find t?

 

[TJGilb]

what is e?

In the case of his equation $U(t)=\frac 1 2 k A^2 sin^2(\omega t+e)$ e is the phase for the harmonic oscillator. It determines the starting point for the sin wave. You can find it using the equation $x=A sin(\omega t+e)$ since you know what x is at time zero.

 

[ikihi]

In the case of his equation $U(t)=\frac 1 2 k A^2 sin^2(\omega t+e)$ e is the phase for the harmonic oscillator. It determines the starting point for the sin wave. You can find it using the equation $x=A sin(\omega t+e)$ since you know what x is at time zero.

Ah I see. So how do you isolate e in this equation: $x=A sin(\omega t+e)$? I don't remember how to do that where the variable is locked up in the sine function.

 

[TJGilb]

Plug in your two known values for position and time. This will get you $25cm=Asin(e)$ since $\omega t$ goes to zero. Knowing how a sin wave behaves, and that 25 centimeters is your max amplitude, you should then be able to solve for e.

 

[ikihi]

Plug in your two known values for position and time. This will get you $25cm=Asin(e)$ since $\omega t$ goes to zero. Knowing how a sin wave behaves, and that 25 centimeters is your max amplitude, you should then be able to solve for e.

well I did (0.25 m / 0.25 m) = sin(e) ----> 1=sin(e)

 

[TJGilb]

well I did 0.25/0.25 = sin(e) ----> 1=sin(e)

Right, now take the inverse sin of both sides and you'll have e.

 

[ikihi]

ok thanks. I forgot about the inverse function.
So e = 1.571

 

[ikihi]

Right, now take the inverse sin of both sides and you'll have e.

So is this the correct plot then?

ω= 2π / T ---> 1.000631908

u(t) = 0.5 * 138 * (0.25)² * sin^2(1.000631908 ⋅ t +1.571)
simplifies to:
u(t) = 4.3125 * sin^2 (1.000631908 ⋅ t +1.571)

 

[TJGilb]

It looks right, except I'm confused about your value for $\omega$. It should be equal to $\frac {2\pi} T$ where T is 1 such that it completes a full cycle every second. For the record, it's also usually easier (and more accurate) to keep things out of decimal form in your equations.

 

[ikihi]

It looks right, except I'm confused about your value for $\omega$. It should be equal to $\frac {2\pi} T$ where T is 1 such that it completes a full cycle every second. For the record, it's also usually easier (and more accurate) to keep things out of decimal form in your equations.

Ah I see. I had an error with my T.
ω should equal 2.51.

Here's the new graph:

 

[TJGilb]

ω should equal 2.51

Still not quite right. Your $\omega$ should equal $2\pi$ if you want it to make one full cycle (from 25cm to -25cm back to 25cm) every second. This isn't a calculated value in this case. The problem statement specifies what they want the period to be.

 

[ikihi]

Still not quite right. Your $\omega$ should equal $2\pi$ if you want it to make one full cycle (from 25cm to -25cm back to 25cm) every second. This isn't a calculated value in this case. The problem statement specifies what they want the period to be.

Okay so the period is 1 rev/sec since it's a clock right?
So ω= 2π/1 = 6.283185307

Here is the graph:

 

[TJGilb]

Precisely.

 

[ikihi]

It should be periodic, since you know the solutions for harmonic oscillators (x=A sin(wt+e)) and the spring potential energy (U(x)=1/2kx^2) and just substitute ,
U(t)=1/2 k A^2 sin^2(wt+e) , now you can plot this function with numerical values.

I'm still confused, how did you get U(t)=1/2 k A^2 sin^2(wt+e). How did that substitution work?

 

[TJGilb]

I'm still confused, how did you get U(t)=1/2 k A^2 sin^2(wt+e). How did that substitution work?

The potential energy of a spring is given by the equation $U(t)=\frac 1 2 kx^2$. He subbed in the equation of x for a harmonic oscillator, $x(t)=Asin(\omega t +\phi)$, into it.