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[cepheid]

to be honest I don't understand what's wrong with them? isn't the double derivative of 3g/2Lcost = −3g/2Lcost

Yes, but this is not equal to -3g/2L times the original function. If the original function is (3g/2L)cos(t), then -3g/2L*(original function) would be -(9g²/4L²)cos(t), which is NOT what you get when you differentiate twice.

So, your proposed function does NOT satisfy your differential equation.

Think about this for a second. What do we know? SOME SORT of sinusoidal function will satisfy the differential equation, but we don't know its exact form. So, in order to figure it out, we need to take into account everything that we DO know:

- we know that at t = 0, angle(t) = A, which means that the function has to be of the type that is non-zero when you give it a zero argument (this helps you decide between sine and cosine)

- we know that A is the maximum displacement, which means that the amplitude of the oscillation is A, which means that the rod goes back and forth between an angular position of +A and -A. This tells you that A has to be multiplying the sinusoidal function in front (and nothing else).

- we know that the frequency has to be in there somewhere and should be set by the physical properties of the rod and its environment. We also know that it should be consistent with the differential equation, which contains that constant C that appears out front AFTER twice differentiating (but maybe now I am giving away too much...)

 

[nabaa]

cosine because at angle = 0 cosine is 1

A in front, okayAcost/(3g/2L)??

??

 

[cepheid]

cosine because at angle = 0 cosine is 1

A in front, okayAcost/(3g/2L)??

??

Mean answer: Why are you asking me?! Differentiate it twice and see if it works or not.

Nice answer: Working "backwards" is difficult, so I will give you a BIG hint. What is the physical significance of the parameter \omega in the expression:

\theta (t) = A\cos(\omega t)

What do you get when you differentiate it once?

\frac{d\theta (t)}{dt} = \frac{d}{dt}[A\cos(\omega t)] =~?

What do you get when you differentiate it again?

\frac{d^2 \theta (t)}{dt^2}= \frac{d^2}{dt^2}[A\cos(\omega t)] =~?

Assuming you STARTED with a function of this form, what must be true in order for the result of the above to match your differential equation?

 

[nabaa]

differentiating once would yield -wAsin(wt)

twice would yield

-w^2Acos(wt)

so w^2 must = 3g/2L

w = sqrt(3g/2L)

function = Acos(sqrt(3g/2L)t), yes?

 

[cepheid]

differentiating once would yield -wAsin(wt)

twice would yield

-w^2Acos(wt)

so w^2 must = 3g/2L

w = sqrt(3g/2L)

function = Acos(sqrt(3g/2L)t), yes?

Beautiful! You have just answered BOTH part (b) and part (c) since T = 1/f and w = 2*pi*f is the angular frequency. Notice how similar the expression for T is to the expression for the period of a *simple* pendulum. The latter is derived in a similar way, by setting up the equation of motion. Now you know where that comes from! It is ridiculously late in my time zone, so, good night!

 

[nabaa]

thank you so much! mine too!