- #1

physicsissohard

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- Homework Statement
- A pendulum clock loses 12s a day if the temperature is 40∘C and gains 4s a day if the temperature is 20∘C . The temperature at which the clock will show correct time, and the coefficient of linear expansion (α) of the metal of the pendulum shaft are respectively.

- Relevant Equations
- delta

I have a problem with the method that they solved. This is what I mean ##\delta t= \frac{\pi L\alpha \delta T}{\sqrt{gL}}##. You can derive this equation by using errors and approximations here delta t is a tiny(not infinitesimal) change in time period, delta T is a tiny change in temperature, alpha is coefficient of linear expansion g and L are gravity and Length.

This equation is used for calculating the change in the time period of the pendulum. Now you need to use the information to set up two equations and solve.

case (i): If change in time period is positive it means the final time period is greater than the accurate time period so pendulm clock loses some amount of time. Now the difference in time period in this case multiplied by numer of oscillations is 12 seconds, because the slower pendulum is behind by some amount for 1 oscillation of the accurate pendulum and if you multiply this difference by number of oscillations in a day it should give 12.

This is how you write 1st equation $12=\frac{43200\pi L\alpha(40-T_a)}{\sqrt{gL}}$ T_a is accurate time T. This all is fine, Now the problem comes in the second case or second equation. They naively constructed the second equation in a very identical manner. instead of 12 they replaced it with -4 and instead of 40 they replaced it with 20 as given in the question. This is where I have a problem Ok whats the meaning of this equation. They basically said that the difference in time period multiplied with number of oscillations gives how much the pendulum clock gained which is 4, and this is wrong. This is wrong interpretation, the difference in time period gives how much the faster pendulum is ahead of the accurate pendulum in *ONE OSCILLATION OF THE FASTER PENDULUM*.

NOte this is not the same as how much faster the pendulum is ahead of the accurate pendulum in one oscillation of the accurate pendulum. So if you multiply this by it by 43200 i.e number of oscillations you get the how much your faster pendulum is ahead of the accurate pendulum not in 1 day but one day according to the faster pendulum. The jee solution formed the solution incorrectly and divided both the equations and obtained temperature as 25. Can someone tell if I made a mistake or Jee question is wrong

This equation is used for calculating the change in the time period of the pendulum. Now you need to use the information to set up two equations and solve.

case (i): If change in time period is positive it means the final time period is greater than the accurate time period so pendulm clock loses some amount of time. Now the difference in time period in this case multiplied by numer of oscillations is 12 seconds, because the slower pendulum is behind by some amount for 1 oscillation of the accurate pendulum and if you multiply this difference by number of oscillations in a day it should give 12.

This is how you write 1st equation $12=\frac{43200\pi L\alpha(40-T_a)}{\sqrt{gL}}$ T_a is accurate time T. This all is fine, Now the problem comes in the second case or second equation. They naively constructed the second equation in a very identical manner. instead of 12 they replaced it with -4 and instead of 40 they replaced it with 20 as given in the question. This is where I have a problem Ok whats the meaning of this equation. They basically said that the difference in time period multiplied with number of oscillations gives how much the pendulum clock gained which is 4, and this is wrong. This is wrong interpretation, the difference in time period gives how much the faster pendulum is ahead of the accurate pendulum in *ONE OSCILLATION OF THE FASTER PENDULUM*.

NOte this is not the same as how much faster the pendulum is ahead of the accurate pendulum in one oscillation of the accurate pendulum. So if you multiply this by it by 43200 i.e number of oscillations you get the how much your faster pendulum is ahead of the accurate pendulum not in 1 day but one day according to the faster pendulum. The jee solution formed the solution incorrectly and divided both the equations and obtained temperature as 25. Can someone tell if I made a mistake or Jee question is wrong

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