# Building a Clock from a Spring: Exploring Potential Energy

• ikihi
In summary, you will need 3.50 kg of mass to make your clock measure seconds. The max velocity of the attached mass is 1.57 m/s. The spring potential energy (Usp) is a min and a max for two periods of the clock.
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.

Last edited:
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.

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?

you can find e using the initial condition

Hiranya Pasan said:
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?

Last edited:
ikihi said:
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
TJGilb said:
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.

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.

TJGilb said:
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)

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• sine.PNG
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ikihi said:
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

TJGilb said:
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)

#### Attachments

• plot.PNG
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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
TJGilb said:
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:

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• plot final.PNG
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ikihi said:
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.

TJGilb said:
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:

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• plot very final.PNG
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Precisely.

ikihi
Hiranya Pasan said:
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?

ikihi said:
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.

## 1. What is potential energy?

Potential energy is the energy that an object possesses due to its position or condition. It is stored energy that has the potential to do work in the future.

## 2. How does a clock work using a spring?

A clock works using a spring by converting the potential energy stored in the spring into kinetic energy as the spring unwinds. This kinetic energy is then used to power the gears and hands of the clock, causing them to move and keep time.

## 3. Can any type of spring be used to build a clock?

No, not all springs are suitable for building a clock. The spring must be able to store and release energy consistently and reliably. Clockmakers typically use spiral or coiled springs made of steel or brass for their mechanical clocks.

## 4. What is the relationship between potential energy and the spring in a clock?

The spring in a clock stores the potential energy that is needed to power the clock. As the spring unwinds, this potential energy is converted into kinetic energy, which is used to keep the clock running.

## 5. How is potential energy calculated in a spring-powered clock?

The potential energy in a spring-powered clock can be calculated using the formula PE = 1/2kx^2, where PE is potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

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