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Tension in a Simple Pendulum

https://isaacphysics.org/questions/tension_pendulum_num?board=0035a9c2-b1ab-4ceb-ac9b-07367ec42a58&stage=a_level

I managed to get the right answer but I did a weird method by solving it as a quadratic.
Could anyone tell me how to do this question the intended way?
Help much appreciated.

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Reply 1
Working:
(edited 2 months ago)
Reply 2
The hint suggests going down the shm (small angle) route so
d^2 theta/dt^2 + w^2 theta = 0
where w^2 = g/l. The solution is fairly easy / well known for the given initial (angular displacement).

Then the tension must be a combination of the (resolved) weight and the centripetal force
(edited 2 months ago)
Reply 3
1) why is "d^2 theta/dt^2 + w^2 theta = 0" - I thought d^2 theta/dt^2 = w^2?
2) When you resolve for tension how do you get it in the form T(t)=T0​−T2​(cos2ωt)+...
Reply 4
Original post by mosaurlodon
1) why is "d^2 theta/dt^2 + w^2 theta = 0" - I thought d^2 theta/dt^2 = w^2?
2) When you resolve for tension how do you get it in the form T(t)=T0​−T2​(cos2ωt)+...
Thats the basic shm ode so
d^2x/dt^2 + w^2x = 0
here x=theta. If you had
d^2x/dt^2 + w^2 = 0
it would integrate up to a quadratic rather than a trig function.

You dont really get that directly, theyve used the small angle approximation for the cos terms then mapped it back to cos (as far as I can see).
Reply 5
so you start off with
d^2 theta/dt^2 + w^2 theta = 0
integrate twice to get theta as a function of time.
Then use w=g/l
and resolve forces?
Reply 6
Original post by mosaurlodon
so you start off with
d^2 theta/dt^2 + w^2 theta = 0
integrate twice to get theta as a function of time.
Then use w=g/l
and resolve forces?
You should know the solution of the shm ode is
theta = theta_0 cos(wt)
but theres no harm in being able to derive it.

Then resolving along the string you have a fairly simple
T = resolved weight + centripetal
and its just thinking about/fiddling with the resolved weight part.
Reply 7
Im confused on what you mean about resolved weight - cant you just plug in the equation for theta in Tsin(theta)=mw^2*l
and sin(theta) will just equal theta or Tcos(theta)=mg and get the answer from there?
Reply 8
Original post by mosaurlodon
Im confused on what you mean about resolved weight - cant you just plug in the equation for theta in Tsin(theta)=mw^2*l
and sin(theta) will just equal theta or Tcos(theta)=mg and get the answer from there?
There are a few ways you could do it and tbh, the question could be a bit clearer about where youre supposed to use the small angle approximations. In your OP, youve got centripetal is horizontal as when you resolve vertically youve ignored it. Similarly for the horizontal force equation, as theta goes to zero (small angle approximation) tension must become infinite and the angularlinear speed is not constant.

To get T, just resolve forces along the string
T = mgcos(theta) + mv^2/l
which is valid for all angles, then use the small angle / shm motion for theta.

Its the standard approach for these sort of pendulum questions. When you get the "right" approximation/solution, why not check what your other approaches give?
(edited 2 months ago)
Reply 9
Oh actually that equation makes a lot more sense than what I was doing.
I didnt consider that tension would go to infinity at theta=0.
Ok so I took that equation and replaced it with mw^2l

But I still cant get it to the T(t)=T0​−T2​(cos2ωt)+... used by isaac :frown:
Reply 10
Original post by mosaurlodon
Oh actually that equation makes a lot more sense than what I was doing.
I didnt consider that tension would go to infinity at theta=0.
Ok so I took that equation and replaced it with mw^2l

But I still cant get it to the T(t)=T0​−T2​(cos2ωt)+... used by isaac :frown:
There are a few mistakes in what you did.

Im not sure what you did to get the angular frequency sqrt(g/l) relationship, but w is not the angular speed as it is in circular motion where the angular speed is constant. Here dtheta/dt varies as its shm. So you can differentiate the theta(t) = ... to get the corresponding angular speed at time t and use it in the centripetal expression.

Then think about the form of the answer they way so
cos(2...) = cos^2 - sin^2
and think about what youre doing the small approximation on.

When youre developing the forces/equations of motion its worth thinking about the forces, expressions, ... when theta=0 and theta=+/-theta_0 etc. When theta=0 here, there is maximum angular speed so the centripetal is maximum and tension will be largest as weight is vertical. Similarly, the centripetal is zero (as the speed is zero) when at +/-theta_0 and the weight component is also minimum so the tension will be minimum. .... Here the centripetal force must be the difference between tension annd the resolved weight as its the radial force acting on the object which ensures circular motion. Similarly, shm arises from the angular acceleration which is driven by resolved weight tangential to the circle. Both of these were wrong in the OP at the start of your working.
(edited 2 months ago)
Reply 11
wait what???
so in shm, the angular frequency (w) is sqrt(g/l) - given by the isaac q, but the angular velocity is completely different.
But in circular motion they are the same - w represents both?
That actually makes sense now that im thinking about it as in shm, frequency is constant but angular velocity is always changing.
(edited 2 months ago)
Reply 12
Original post by mosaurlodon
wait what???
so in shm, the angular frequency (w) is sqrt(g/l) - given by the isaac q, but the angular velocity is completely different.
But in circular motion they are the same - w represents both?
That actually makes sense now that im thinking about it as in shm, frequency is constant but angular velocity is always changing.
Thats correct. Id guess your original derivation of w=sqrt(g/l) should have highlighted this. The w^2 here is the term in the shm equation so
d^2theta/dt + w^2 theta = 0
Its not dtheta/dt which is the angular velocity which must vary between 0 at the extremes +/-theta_0 and max/min at theta=0. The centripetal force at time t depends on the angular (or linear) speed at that instant.
Reply 13
Original post by mosaurlodon
wait what???
so in shm, the angular frequency (w) is sqrt(g/l) - given by the isaac q, but the angular velocity is completely different.
But in circular motion they are the same - w represents both?
That actually makes sense now that im thinking about it as in shm, frequency is constant but angular velocity is always changing.
The hint about
cos(2...) = cos^2 - sin^2
so think about whether you need to do a small angle approximation for the centripetal (or not, hint).
Reply 14
Thank you for being so patient :smile:. I owe you plentiful gratitude
Im so close I can feel it

The terms are starting to make sense
(edited 2 months ago)
Reply 15
Original post by mosaurlodon
Thank you for being so patient :smile:. I owe you plentiful gratitude
Im so close I can feel it

The terms are starting to make sense
Getting there, on the T= line, youve written angular acceleration (second derivative) whereas its the first derivative (angular velocity) squared. The next lline is ok, so its what you intended, but not what youve written.

Looks like youve got an extra 1/2 otherwise youd be there. Id just been doing it in my head, so will have a quick scribble.
Reply 16
Original post by mosaurlodon
Thank you for being so patient :smile:. I owe you plentiful gratitude
Im so close I can feel it

The terms are starting to make sense
When I did it in my head, I thought the cos^2 and sin^2 terms would just combine, but there are 3 formula for cos(2...) and the pythagoras identity on sin^2 (or cos^2, you choose) should get you there.
(edited 2 months ago)
Reply 17

so like this? - this answer does get me 74.
Reply 18
Original post by mosaurlodon

so like this?.
Not checked it/worked it through but it looks about right (and stealing your answer gives isaac correct).
(edited 2 months ago)
Reply 19
Original post by mosaurlodon

so like this? - this answer does get me 74.
Id probably work through

Constant speed circular motion, starting with x = (cos,sin) and deriving the speed and radial centripetal force

Make sure you did the first part of this question where w=sqrt(g/l)

Make sure you sketch/write down the forces for this question

Be clear about how the small angle parts are used for shm

Think about working back from the answer here so cos(2...) = ... and use that to guide your working.

Anyway off to see the Pixies now so keep busy

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