Is the Classical Pendulum Formula Still Accurate for Large Theta Values?

[rehab]

Summary: Hi, I'm trying to solve this problem, if it's not right then please help me with a hint without solving it.

This formula is just an approximation for small values of theta, but if Vo was greater than the denominator this will lead to large values of theta and then this solution is not valid anymore, I'm not sure so I need your

 

[kuruman]

Your expression for $\theta(t)$ does not match the statement of the problem because $\theta(0) \neq 0$. If I were doing this, I would come up with a ceiling for "small angle", e.g. $\theta \leq f\pi$ where $f<1$. Then I would figure out the ceiling for $v_0$.

 

[rehab]

Your expression for $\theta(t)$ does not match the statement of the problem because $\theta(0) \neq 0$.

Why?
the problem stated that at first, it was at rest in equilibrium!

 

[kuruman]

Why?
the problem stated that at first, it was at rest in equilibrium!

What is $\cos0^o$?

 

[rehab]

What is $\cos0^o$?

1.
but it doesn't add anything because the solution theta= A cos(theta) 'let's ignore the sin term' at t =0 is
theta = A and from the problem, this is just zero.
Right?

 

[haruspex]

1.
but it doesn't add anything because the solution theta= A cos(theta) 'let's ignore the sin term' at t =0 is
theta = A and from the problem, this is just zero.
Right?

I assume you mean $\theta=A\cos(\omega t)$, but as @kuruman pointed out, that violates $\theta(0)=0$. So what equation should you write instead?

if Vo was greater than the denominator this will lead to large values of theta

How large, exactly, and how large a value would render the solution invalid?

 

[kuruman]

1.
but it doesn't add anything because the solution theta= A cos(theta) 'let's ignore the sin term' at t =0 is
theta = A and from the problem, this is just zero.
Right?

You wrote down
$\theta(t)=\dfrac{\dot{\theta}_0}{\Omega}\cos(\Omega~t).$
At $t=0$,
$\theta(0)=\dfrac{\dot{\theta}_0}{\Omega}\cos(\Omega*0)=\dfrac{\dot{\theta}_0}{\Omega}\cos(0)=\dfrac{\dot{\theta}_0}{\Omega}*(1) =\dfrac{\dot{\theta}_0}{\Omega}.$

The result is not zero.

 

[rehab]

You wrote down
$\theta(t)=\dfrac{\dot{\theta}_0}{\Omega}\cos(\Omega~t).$
At $t=0$,
$\theta(0)=\dfrac{\dot{\theta}_0}{\Omega}\cos(\Omega*0)=\dfrac{\dot{\theta}_0}{\Omega}\cos(0)=\dfrac{\dot{\theta}_0}{\Omega}*(1) =\dfrac{\dot{\theta}_0}{\Omega}.$

The result is not zero.

Oh, yeah I just realized that I inadvertently wrote down the cos function where it should be the sin function instead, is it right now?
And what do think about this expression for Vo?

 

[rehab]

I assume you mean $\theta=A\cos(\omega t)$, but as @kuruman pointed out, that violates $\theta(0)=0$. So what equation should you write instead?

How large, exactly, and how large a value would render the solution invalid?

It should be written with the sic function for some reason I wrote it with the cos function.

for Vo:

 

[kuruman]

And what do think about this expression for Vo?

Not much. Your expression has time parameter $t$ in it which doesn't make sense. You should find an expression that one can use in an actual pendulum experiment. In other words, suppose I gave you a real pendulum, with length $L$ that you measure and find it is 1.5 m. How would you proceed to find a numerical answer for $v_0$ in m/s? Show a sample calculation.

 

[rehab]

Not much. Your expression has time parameter $t$ in it which doesn't make sense. You should find an expression that one can use in an actual pendulum experiment. In other words, suppose I gave you a real pendulum, with length $L$ that you measure and find it is 1.5 m. How would you proceed to find a numerical answer for $v_0$ in m/s? Show a sample calculation.

well, simply measure the initial {\theta} and then {\theta} at some stop time $t$, then multiply the result with $L$.
why does having $t$ as a parameter doesn't make sense?

 

[haruspex]

why does having t as a parameter doesn't make sense?

Because you want the range of v₀ for which a function of t is valid in general, not for some particular values of t.

 

[rehab]

Because you want the range of v₀ for which a function of t is valid in general, not for some particular values of t.

So, should I replace $t$ by theta/$L$?

 

[kuruman]

So, should I replace $t$ by theta/$L$?

Why would you do that? That relation is incorrect. Elapsed time is not angle divided by pendulum length. I suggest that you start by finding a relation between $v_0$ and the maximum angle reached by the pendulum.

 

[rehab]

Why would you do that? That relation is incorrect. Elapsed time is not angle divided by pendulum length. I suggest that you start by finding a relation between $v_0$ and the maximum angle reached by the pendulum.

It's 12:11 here and this is time for bed this is clearly the reason for this unforgivable mistake :).
well, I can't think of anything but (f*pi''maximum angle''/2*pi ''the total period'') *L.

 

[kuruman]

Please get some rest, clear your head and come back later. We'll be here.

 

[haruspex]

This formula is just an approximation for small values of theta

Right, so what is the exact formula? Can you estimate the difference?