- 13 Oct '08 22:18A mass (
*m*) is hanging on a massless, unstretchable rod of length*l*, and it swings like a pendulum. There is a frictinal force proportional to the velocity of the mass.

a) What is the equation of motion for the mass?

b) In the approximation of small ampiltudes (sin x = x), show that the natural frequency is (g/*l*)^(1/2), where g is the gravitational field strength.

c) Neglecting friction, but considering large amplitudes ( sin x =/= x), determine the period*T*. - 15 Oct '08 15:34 / 1 edit

Missed this one! Now let's see...*Originally posted by Jirakon***A mass (***m*) is hanging on a massless, unstretchable rod of length*l*, and it swings like a pendulum. There is a frictinal force proportional to the velocity of the mass.

a) What is the equation of motion for the mass?

b) In the approximation of small ampiltudes (sin x = x), show that the natural frequency is (g/*l*)^(1/2), where g is the ...[text shortened]... ecting friction, but considering large amplitudes ( sin x =/= x), determine the period*T*.

**(a) What is the equation of motion for the mass?**

The arc length "s" swept out through an angle "theta" by a pendulum of length "L" is given by:

s = L * theta

To find the velocity, we differentiate with respect to time:

v = ds/dt = L * dtheta/dt

The friction force is proportional to the velocity, so we have:

Ff = kL * dtheta/dt

Differentiating "v" with respect to time we get:

a = dv/dt = L * d^2theta/dt^2

Multiplying this by the mass gives us the net force:

F = ma = mL * d^2theta/dt^2

Also, a force balance on a free-swing mass attached to the pendulum that is constrained to swing on a circular arc reveals that the resultant force is:

F = -mg * sin(theta)

This value is negative, because a positive displacement results in a force back along the arc. Summing this force and the friction force gives us:

F(total) = kL * dtheta/dt - mg * sin(theta)

Equating this expression with our other force expression, we get:

mL * d^2theta/dt^2 = kL * dtheta/dt - mg * sin(theta)

Simplifying, we have:

d^2theta/dt^2 - (k/m) * dtheta/dt + (g/L) * sin(theta) = 0

This is the expression for the motion of the pendulum when the friction force is proportional to the velocity. I think it can be solved using Laplace transforms, but I haven't done them in a while so I'll leave them for another day.

**(b) In the approximation of small ampiltudes (sin x = x), show that the natural frequency is (g/l)^(1/2), where g is the gravitational field strength.**

In approximating small amplitudes, I'll make the assumption that the velocity is small as well and so we can neglect friction. Therefore, our expression above reduces to:

d^2theta/dt^2 + (g/L) * theta = 0

Using the D-operator method to solve this, we get:

(D^2 + (g/L))(theta) = 0

D = +/- i*(g/L)^0.5

We could plug this value into the characteristic expression, but all we're really interested in is the magnitude of the constant, which denotes the frequency (g/L)^0.5 as required.

**(c) Neglecting friction, but considering large amplitudes ( sin x =/= x), determine the period T.**

For large amplitudes with no friction, our expression reduces to:

d^2theta/dt^2 + (g/L) * sin(theta) = 0

According to Wikipedia, after solving this expression for dtheta/dt and then inverting, the result can't be integrated so you can only approximate the answer using a power series.

http://en.wikipedia.org/wiki/Pendulum_(mathematics)#Arbitrary-amplitude_period - 05 Nov '08 11:18

PBE6 i always love reading your posts. even though i've never taken any serious physics, the running commentary always carries the reader right through the steps in a logical and concise manner. just wanted to thank you on behalf of the math/physics nerd community*Originally posted by PBE6***Missed this one! Now let's see...**

[b](a) What is the equation of motion for the mass?

The arc length "s" swept out through an angle "theta" by a pendulum of length "L" is given by:

s = L * theta

To find the velocity, we differentiate with respect to time:

v = ds/dt = L * dtheta/dt

The friction force is proportional to the velocity, so we ...[text shortened]...

http://en.wikipedia.org/wiki/Pendulum_(mathematics)#Arbitrary-amplitude_period[/b] - 05 Nov '08 15:44

It all comes from old engineering test habits - going for part marks!!*Originally posted by Aetherael***PBE6 i always love reading your posts. even though i've never taken any serious physics, the running commentary always carries the reader right through the steps in a logical and concise manner. just wanted to thank you on behalf of the math/physics nerd community**

Thanks for the kind words, though. I always strive for clarity and concision when writing a solution, for my own benefit as much as anyone else's. I find that if I can't write a solution out in words, then I haven't understood the ideas. I'm glad to hear this ensures my ideas are clearly translated to the reader (right or wrong ). - 07 Nov '08 05:18

I second Aetherael about your posts up in tha P&P!*Originally posted by PBE6***It all comes from old engineering test habits - going for part marks!!**

Thanks for the kind words, though. I always strive for clarity and concision when writing a solution, for my own benefit as much as anyone else's. I find that if I can't write a solution out in words, then I haven't understood the ideas. I'm glad to hear this ensures my ideas are clearly translated to the reader (right or wrong ).