Some question from the oxford entrance test

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Homework Help Overview

The discussion revolves around the behavior of pendulums, specifically focusing on the mathematical treatment of ideal and real pendulums, as well as the effects of temperature on pendulum accuracy. Participants are exploring the implications of given equations and concepts related to pendulum motion and thermal expansion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equation for the period of a real pendulum and the need to determine the effective length and moment of inertia. Questions arise about the implications of limits as certain parameters approach zero. There is also exploration of the relationship between temperature change and pendulum accuracy.

Discussion Status

Some participants are attempting to clarify their understanding of the problem and the mathematical relationships involved. Guidance has been offered regarding the connection between temperature changes and the period of the pendulum, as well as the implications of limits in the context of the problem.

Contextual Notes

Participants are working within the constraints of a homework assignment, which includes specific questions about the behavior of pendulums under varying conditions and the effects of thermal expansion on accuracy. There is a focus on understanding the mathematical relationships without providing direct solutions.

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14. You should already be familiar with the mathematical treatment of an
ideal pendulum, in which the pendulum bob is modeled as a point
mass on the end of a rigid rod of negligible mass. In this problem you
will consider the behaviour of more complex types of pendulum. You
will be given all the information you need in the sections below.
For a general pendulum of any shape and size the period P is given by
p= 2*pie*root(I/gM(LCM))
where g is the acceleration due to gravity, M is the total mass of the
pendulum, LCM is the effective length of the pendulum, defined as the
distance from the pivot to the centre of mass, and I is the moment of
inertia around the pivot point. For a point mass m fixed at a distance
r from the pivot I = mr^2, while for a uniform rod of mass m and length
r attached to the pivot at one end I = 1/3mr^2
. For more complex objects
the total moment of inertia can be calculated by adding together values
for the component parts.

(c) Now consider the case of a real pendulum, with a bob of mass
Mb (which you may treat as a point mass) attached to the pivot
using a uniform rod of mass Mr and length L, and find the period
in this case. Show that the result for a real pendulum reduces to
the results for an ideal pendulum and a rod pendulum by taking
appropriate limits.
 
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What have you tried? You are given the equation for the period of the real pendulum...all you need to do is determine LCM and I. What happens to p as Mr approaches zero?
 
okokok thank you dat s the direction i was thinking but i don't kno whether it s the right one

(d) Most substances expand with increasing temperature, and so a
metal rod will expand in length by a fraction ®dT if the tem-
perature is changed by dT, where ® is called the coe±cient of
linear thermal expansion. Consider the effect of this expansion
on a pendulum clock with a pendulum made from brass, with
® = 19 £ 10¡6 K¡1. What temperature change can this clock
tolerate if it is to remain accurate to 1 second in 24 hours? [5]

do u just try to separate p and dp and stuff??
 
wat is the question saying wen it is asking u to take appriopriate limit??
 
i25959341 said:
wat is the question saying wen it is asking u to take appriopriate limit??

For the ideal pendulum, Mr=0 so the "appropriate limit is the limit as Mr approaches zero.
 
i25959341 said:
okokok thank you dat s the direction i was thinking but i don't kno whether it s the right one

(d) Most substances expand with increasing temperature, and so a
metal rod will expand in length by a fraction ®dT if the tem-
perature is changed by dT, where ® is called the coe±cient of
linear thermal expansion. Consider the effect of this expansion
on a pendulum clock with a pendulum made from brass, with
® = 19 £ 10¡6 K¡1. What temperature change can this clock
tolerate if it is to remain accurate to 1 second in 24 hours? [5]

do u just try to separate p and dp and stuff??
the period p will be a function of the length of the pendulum and the length of the pendulum will be a function of temperature...therefor p is a function of temperature... if the temperature change is [itex]\Delta T=T_2-T_1[/itex] what is the change in period [itex]\Delta p = p(T_2)-p(T_1)[/itex]? If the pendulum is accurate to within 1 second in 24 hours, then [itex]\Delta p \leq \frac{1s}{24h}[/itex] and you can solve for [itex]\Delta T[/itex].
 
thabk you
 
24. A black box has three electrical connections labelled A, B, and C,
arranged in a triangle as shown below.

A
b c
The box contains three components: a resistor, a small capacitor and a
diode. You know that one component is connected between each pair
of terminals, but you cannot see exactly how they are arranged. You
make the following observations with a 9V battery connected in series
with an ammeter
² When the battery is connected with + to A and ¡ to B a current
of 3mA °ows
² When the battery is connected with + to B and ¡ to C a very
large current °ows
² When the battery is connected with + to C and ¡ to A no current
is measured
² When the battery is connected with ¡ to B and + to C no current
is measured
 
i25959341 said:
wat is the question saying wen it is asking u to take appriopriate limit??

Will you be tested on "ure" grammar?
 
  • #10
gabbagabbahey said:
the period p will be a function of the length of the pendulum and the length of the pendulum will be a function of temperature...therefor p is a function of temperature... if the temperature change is [itex]\Delta T=T_2-T_1[/itex] what is the change in period [itex]\Delta p = p(T_2)-p(T_1)[/itex]? If the pendulum is accurate to within 1 second in 24 hours, then [itex]\Delta p \leq \frac{1s}{24h}[/itex] and you can solve for [itex]\Delta T[/itex].

OK so I got to √L2 - √L1 ≤ 5.825 x 10^-6

How do I get ΔL from there?
 
  • #11
You got Δ√L haven't you?
 
  • #12
Yeah, but how can I get ∂T from ∆√L ?
 
  • #13
You are looking for the maximum change in length allowed aren't you? So if you have the square root of the change in length allowed..
 
  • #14
I'm sorry, I still don't understand, if I had √∆L then I could do it, but that's not equal to ∆√L is it?
 

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