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.
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.
Scroll to see replies
Reply 1
10
Working:
(edited 1 year ago)
Reply 2
21
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
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 1 year ago)
Reply 3
10
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)+...
2) When you resolve for tension how do you get it in the form T(t)=T0−T2(cos2ωt)+...
Reply 4
21
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)+...
2) When you resolve for tension how do you get it in the form T(t)=T0−T2(cos2ωt)+...
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
10
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?
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
21
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?
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?
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
10
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?
and sin(theta) will just equal theta or Tcos(theta)=mg and get the answer from there?
Reply 8
21
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?
and sin(theta) will just equal theta or Tcos(theta)=mg and get the answer from there?
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 1 year ago)
Reply 9
10
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
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
Reply 10
21
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
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
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 1 year ago)
Reply 11
10
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.
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 1 year ago)
Reply 12
21
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.
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.
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
21
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.
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.
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
10
Thank you for being so patient 
. I owe you plentiful gratitude
Im so close I can feel it
The terms are starting to make sense
. I owe you plentiful gratitudeIm so close I can feel it
The terms are starting to make sense
(edited 1 year ago)
Reply 15
21
Original post by mosaurlodon
Thank you for being so patient . I owe you plentiful gratitude
Im so close I can feel it
The terms are starting to make sense
. I owe you plentiful gratitudeIm so close I can feel it
The terms are starting to make sense
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
21
Original post by mosaurlodon
Thank you for being so patient . I owe you plentiful gratitude
Im so close I can feel it
The terms are starting to make sense
. I owe you plentiful gratitudeIm so close I can feel it
The terms are starting to make sense
(edited 1 year ago)
Reply 17
10
so like this? - this answer does get me 74.
Reply 18
21
Original post by mosaurlodon
so like this?.
(edited 1 year ago)
Reply 19
21
Original post by mosaurlodon
so like this? - this answer does get me 74.
•
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.
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