Seeking advice on "simple" pendulum problem

[some bloke]

[Mentors' note: moved here from the technical forums after the thread was already fully developed, so no HW template]

Hello all!

I have a problem which I am beginning to suspect I am either unequipped to solve, or which does not have enough details! I can see, in my mind, the problem, but I cannot work out how to bring the logic backwards to a conclusion!

I have an undamped simple pendulum of length 0.8m, which I know is horizontally displaced by 0.022m when its velocity towards the equilibrium is 1.4m/s.

I need to work out the Amplitude and the period of the pendulum.

I have worked out the natural frequency is 3.5, and the time period is 1.79. That bit was fine. But with this information, I cannot work out how to calculate the Amplitude! I have gone in circles by making symultaneous equations using s=A cos(ωt) and v=-A ωsin(ωt), but I ended up with an amplitude of 45m, which is obviously ridiculous!

My Google-Fu is not working well for me and I feel like I'm missing something. I managed to get a value of t=0.44s, so the pendulum is at position s=0.022m when t=0.44s, but I after some bungling of the next steps I ran an excel "goal seek" to change t to make the Amplitude values from reversing both the displacement and velocity formulae and it gave a value of around 0.433s, and this made me notice there was a huge change in Amplitude when the time was changed just a tiny bit.

I'm very much lost on this one, but I don't want to give up - I just need a nudge in the right direction, chances are I'm overcomplicating this a lot and it's something simple I've overlooked!

Thanks in advance!

 

[Ibix]

Hint: $\cos^2(x)+\sin^2(x)=1$.

 

[Baluncore]

You did not mention energy. Always follow the energy.
The sum of the PE and the KE is constant for an undamped pendulum.
Solve for x and y, when velocity equals zero.

 

[some bloke]

Thanks all! I used the energy side to work it out, I ended up with an Amplitude of 0.388m and a Time period of 1.79s - it all seems reasonable, does this look correct?

 

[Ibix]

I agree the period, but the amplitude looks a bit off. With $g=10\mathrm{ms^{-2}}$ I get the amplitude to be $0.396\mathrm{m}$. How did you get your answer?

 

[some bloke]

I agree the period, but the amplitude looks a bit off. With $g=10\mathrm{ms^{-2}}$ I get the amplitude to be $0.396\mathrm{m}$. How did you get your answer?

That might be as I'm using g=9.81ms^-2, so that might be where the slight difference is?

 

[PeroK]

How is the amplitude of a simple pendulum defined?

 

[some bloke]

How is the amplitude of a simple pendulum defined?

It's the maximum displacement of the pendulum from the equilibrium point?

 

[PeroK]

It's the maximum displacement of the pendulum from the equilibrium point?

How is that calculated?

 

[PeroK]

I agree the period, but the amplitude looks a bit off. With $g=10\mathrm{ms^{-2}}$ I get the amplitude to be $0.396\mathrm{m}$. How did you get your answer?

I don't get the same answer. PS I was using $g = 9.8m/s^2$.

 

[PeroK]

What do you mean by amplitude? The maximum horizontal displacement or the maximum angle?

Both answers were given in metres, so I assumed it was the distance the bob travels from the lowest point to the highest point. However, this distance is only simple harmonic for small angles where $\theta \approx \sin \theta$. That's why we need to define how this distance is calculated.

I'd previous assumed that the amplitude referred to the maximum angle.

 

[some bloke]

The problem frames it that the pendulum is displaced horizontally by 0.022m when the bob is moving at 1.4ms^-1. It then gives you the length of 0.8m and asks for the Amplitude and Period.

Period is 2*pi*root(length/gravitational constant, so that's simple enough to calculate.

Amplitude I have worked out by calculating the total energy based on elevation of the pendulum at the point s=0.022 and v=1.4, using trigonometry, and then working out the maximm elevation where all energy is potential, then using trigonometry to calculate the horizontal displacement, which came to 0.388m.

I will be honest, I would not put it past my tutor to have made the numbers up without checking whether this remains a simple pendulum!

 

[PeroK]

The problem frames it that the pendulum is displaced horizontally by 0.022m when the bob is moving at 1.4ms^-1. It then gives you the length of 0.8m and asks for the Amplitude and Period.

I missed that from the original problem. I used 1.4 m/s at the lowest point. 0.022m is a small displacement compared to the length of the pendulum. It's not going to make much difference, given the pendulum will have lost very little KE during the initial displacement.

Also, 1.4 m/s is quite fast for a pendulum of length only 0.8m. That means that the small angle becomes a relatively poor approximation.

Period is 2*pi*root(length/gravitational constant, so that's simple enough to calculate.

The period of a simple pendulum is independent of the amplitude for small oscillations. Approximately $1.8s$ is the best you can do.

Amplitude I have worked out by calculating the total energy based on elevation of the pendulum at the point s=0.022 and v=1.4, using trigonometry, and then working out the maximm elevation where all energy is potential, then using trigonometry to calculate the horizontal displacement, which came to 0.388m.

I get significantly more than that.

I will be honest, I would not put it past my tutor to have made the numbers up without checking whether this remains a simple pendulum!

It looks like a question that hasn't been thought through. It's absurd to give a displacement to the nearest millimeter and then apply the small angle approximation.

 

[some bloke]

Doing a quick adjustment (I am using excel to work it all out so changing variables is easy) I suspect he changed it from 0.4m/s to 1.4m/s. 0.4 comes out with an amplitude of about 100mm, which is much more like a simple pendulum!

On getting significantly more on amplitude, my first calculations said 45m! which is obviously absurd for a 0.8m long pendulum!

I will be seeing him on Tuesday and there's at least 3 of us who are baffled by it, so we'll have to see what he says!

 

[PeroK]

Doing a quick adjustment (I am using excel to work it all out so changing variables is easy) I suspect he changed it from 0.4m/s to 1.4m/s. 0.4 comes out with an amplitude of about 100mm, which is much more like a simple pendulum!

The reduction in speed would make sense.

 

[PeroK]

Let's do a calculation of the height, $h_0$, for a horizontal displacement, $d_0 = 0.022m$ for a pendulum of length $l = 0.8m$.
$$\theta_0 \approx \sin \theta_0 = \frac {d_0} l = 0.0275$$$$h_0 = l(1 - \cos \theta_0) = 0.0003m = 0.3mm$$You can see that this initial displacement is not going to significantly affect the calculation - compared to taking $d_0 = 0$ for the same initial speed. Let's check by calculating $v$ at the lowest point, assuming the speed is $v_0 = 1.4m/s$, at horizontal displacement $d_0$:
$$v^2 = v_0^2 + 2gh_0$$$$v = 1.402m/s \approx 1.4 m/s$$And we see that this complication makes very little, if any, difference to the calculation.

 

[PeroK]

PS note that you can do my steps in reverse to get the maximum height, angle and horizontal displacement for the original problem.

 

[some bloke]

Ok, I agree on the height at S=0.022 as 0.0003m.

From that I worked out that a constant equal to the total energy multiplied by mass was equal to 1/2v^2 plus gh, which gave a value of 0.983*m.

Then to get maximum height (where Ek is 0) I have mgh=0.983m, thus gh=0.983, so h=0.983/9.81, which came to 0.10m as a maximum height.

Then used trigonometry to get Smax, or Amplitude, equal to 0.388m.

Can't see how you could have an amplitude of 1.4m on a pendulum of 0.8m!

Can't say I know where your equations ov v^2 are coming from, nor how v=1.4 makes s=1.4, as v is a velocity?

 

[PeroK]

Ok, I agree on the height at S=0.022 as 0.0003m.

From that I worked out that a constant equal to the total energy multiplied by mass was equal to 1/2v^2 plus gh, which gave a value of 0.983*m.

Then to get maximum height (where Ek is 0) I have mgh=0.983m, thus gh=0.983, so h=0.983/9.81, which came to 0.10m as a maximum height.

Okay. $h = 0.10m$ looks right.

Then used trigonometry to get Smax, or Amplitude, equal to 0.388m.

Okay, but that needs to be rounded off. It can't be correct to the nearest millimeter.

Can't say I know where your equations ov v^2 are coming from,

Conservation of energy.

nor how v=1.4 makes s=1.4, as v is a velocity?

I'm not sure where you are getting that from.

 

[haruspex]

nor how v=1.4 makes s=1.4, as v is a velocity?

Are you misreading this:

$$v = 1.402m/s \approx 1.4 m/s$$

which means

$$v = 1.402m/s $$$$\approx 1.4 m/s$$

not

$$s \approx 1.4 m/s$$

?

 

[bob012345]

Just a couple of observations. This problem can be done without any energy considerations and the amplitude gives a maximum angle around 28.5° whereas the usual small angle limit is around $\frac{\pi}{8}$ or 22.5° but the errors are still small being on the order of 1% or 0.02s in the period. [corrected angle]

 

[Ibix]

I suspect I did the same calculation as @bob012345 - use the displacement and velocity formulae from the OP with the trigonometric identity in #2.

I observe that my answer is the amplitude of the arc length of the swing, though, and I think OP's is the amplitude of the horizontal displacement. The difference illustrates that the oscillation is not quite "small".

 

[PeroK]

I suspect I did the same calculation as @bob012345 - use the displacement and velocity formulae from the OP with the trigonometric identity in #2.

I observe that my answer is the amplitude of the arc length of the swing, though, and I think OP's is the amplitude of the horizontal displacement. The difference illustrates that the oscillation is not quite "small".

Yes. It seems that $h = 0.10m$, $\theta_{max} = 0.51 \equiv 29.0^{\circ}$, hence the maximum horizontal displacement is $0.39m$, which equates to an arc length of $0.40m$.

 

[some bloke]

Thank you all so much, I feel confident that I've got this one right, though I will still be checking with my tutor tomorrow as I'm certain this was not covered in our lessons, and I want to make sure that I'm not expected to demonstrate a different method!

 

[bob012345]

I suspect I did the same calculation as @bob012345 - use the displacement and velocity formulae from the OP with the trigonometric identity in #2.

I just used the basic relations $\theta=\theta_{max}cos(\omega t)$ and $\dot\theta=\omega\theta_{max} sin(\omega t)$where $\theta=\theta_{max}$ at t=0 and $\theta=\theta_i, \dot\theta=\dot\theta_i$ and $t=t_i$ at the given conditions. Then dividing these to eliminate $\theta_{max}$ I solved for $t_i$

$$t_i= \frac{1}{\omega} tan^{-1}\left(\frac{\dot\theta_i}{\omega\theta_i}\right)$$ then plug back in to solve for $\theta_{max}$ and $x_{max}$.

 

[PeroK]

I just used the basic relations $\theta=\theta_{max}cos(\omega t)$ and $\dot\theta=\omega\theta_{max} sin(\omega t)$

That should have a negative sign.

where $\theta=\theta_{max}$ at t=0 and $\theta=\theta_i, \dot\theta=\dot\theta_i$ and $t=t_i$ at the given conditions. Then dividing these to eliminate $\theta_{max}$ I solved for $t_i$

$$t_i= \frac{1}{\omega} tan^{-1}\left(\frac{\dot\theta_i}{\omega\theta_i}\right)$$ then plug back in to solve for $\theta_{max}$ and $x_{max}$.

Simpler is to use the basic equations to get:
$$\theta_{max}^2 = \theta^2 + \frac{\dot \theta^2}{w^2} = \theta^2 + \frac{v^2}{l^2w^2}$$Normally we have $v$ at $\theta = 0$ (and that's still a good approximation here). In any case, as above, this gives
$$\theta_{max} = 0.5, \ x_{max} = 0.38m$$This also confirms that using $w = \sqrt{\dfrac g L}$ is a good approximation for the angular frequency in this case.

 

[PeroK]

Post script: in the general case where we are given $v$ at $\theta = 0$, we can compare the two methods above.

Using conservation of energy gives the exact answer of $$\cos \theta_{max} = 1 - \frac{v^2}{2gl}$$Using the small angle approximation for SHM with angular frequency $w = \sqrt{\dfrac g L}$ gives$$\theta_{max} = \frac{v}{\sqrt{gl}}$$$$\cos \theta_{max} = 1 - \frac{\theta_{max}^2}{2!} + \frac{\theta_{max}^4}{4!} - \dots$$$$= 1 - \frac{v^2}{2gl} + \frac{v^4}{16g^2l^2} - \dots$$And we see that the approximation is good as long as $\dfrac{\theta_{max}^4}{4!}$ is small. Which is the case here.

 

[bob012345]

That should have a negative sign.

Which I knew would be cancelled out by the explicit sign of $\dot\theta_i$ but you are correct and I should have said something.