JEE solution wrong? (tempco of a clock)

[physicsissohard]

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

 

[haruspex]

For now, just fixing up your LaTeX. (The dollar signs need to be doubled up.)

$$\delta t= \frac{\pi L\alpha \delta T}{\sqrt{gL}}$$

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 replaaced 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

 

[haruspex]

What does it mean to say a clock gains 4s a day? Does it show 24h as having elapsed 4s too soon, or is it that when 24h has passed it is 4s fast? Does it matter?
With one interpretation both equations are strictly correct; with the other, they are near enough, surely? (I don't know why you agreed with one but not the other.)

Btw, there is no need to use differentiation. Just write a few simultaneous equations and solve.

 

[BvU]

Hello and happy 2024 !

Can someone tell if I made a mistake

Since you ask : I suspect you are making up some things. E.g.:

43200 i.e number of oscillations

which isn't given at all -- unless you gave us an incomplete problem statement.
(quite possible, because without some extra info, only $\alpha\over L$ can be determined)

All you have is the two observations (with rather finite precision) that can easily be plotted:

and the 25$^\circ$ pops out immediately. The second part of the exercise involves finding the relationship between $\alpha$ and the slope of the plot.

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

Within the precision of the given 4 and 12 s it definitely is the same.

$\$

 

[physicsissohard]

43200 is the number of oscillation that a pendulum does in a day i.e 24 hours

 

[physicsissohard]

"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 you are saying This statement is factually correct but the compuatational difference between both the interpretations is negligible?

 

[BvU]

43200 is the number of oscillation that a pendulum does in a day i.e 24 hours

Only a pendulum with a 2 second period (so about 1 m long). Is that a given ?

$\$

 

[physicsissohard]

What does it mean to say a clock gains 4s a day? Does it show 24h as having elapsed 4s too soon, or is it that when 24h has passed it is 4s fast? Does it matter?
With one interpretation both equations are strictly correct; with the other, they are near enough, surely? (I don't know why you agreed with one but not the other.)

Btw, there is no need to use differentiation. Just write a few simultaneous equations and solve.

actually for the jee answer to be correct this should be the interpretation of gaining time. which is slight different for the interpretation of losing time. losing seconds means the actual seconds(not slow pendulum seconds) that are needed for the slower pendulum to strike the 24th hour right after the accurate pendulum strikes it. And now for gaining seconds this is the meaning for answer to be correct, The number of seconds(real seconds again) it takes for the accurate pendulum to strike the 24th hour right after the faster pendulum struck the 24th hour.

 

[physicsissohard]

For now, just fixing up your LaTeX. (The dollar signs need to be doubled up.)

I think I figured it out, its a definition problem. Actually for the jee answer to be correct this should be the interpretation of gaining time. which is slightly different from the interpretation of losing time. losing seconds means the actual seconds(not slow pendulum seconds) that are needed for the slower pendulum to strike the 24th hour right after the accurate pendulum strikes it. And now for gaining seconds, this is the meaning for the answer to be correct, The number of seconds(real seconds again) it takes for the accurate pendulum to strike the 24th hour right after the faster pendulum struck the 24th hour.

 

[kuruman]

You haven't answered the question in post #7 by @BvU. Do you really believe that

43200 is the number of oscillation that a pendulum does in a day i.e 24 hours

?
If that is the case, then it is unlikely that you figured it out. Please post your solution and we will confirm its correctness.

 

[Chestermiller]

If it loses 12 seconds a day, it means that the pendulum beats 43200-6 cycles in a day. So, if it loses 12 seconds a day, the period of the pendulum is $86400/(43200-6)=2(1+\frac{6}{43200})=2(1+0.000139)\ sec$. Similarly, if it gains 4 seconds a day, the serious of the pendulum is $2(1-0.0000463)\ sec.$

 

[kuruman]

If it loses 12 seconds a day, it means that the pendulum beats 43200-6 cycles in a day. So, if it loses 12 seconds a day, the period of the pendulum is $86400/(43200-6)=2(1+\frac{6}{43200})=2(1-0.000139)\ sec$. Similarly, if it gains 4 seconds a day, the serious of the pendulum is $2(1+0.0000463)\ sec.$

I am not so sure about this. Imagine a clock that displays time in seconds over a 24-hour period, from 0 to 86,399. A clock that loses 12 seconds over a 24-hour period will display 86,387 when the accurate clock displays 86,399. For that to happen, the period of the "slow" pendulum that regulates its timing must be longer, not shorter, than that of the "good" clock. In other words one must add $\frac{12}{86,400}$ seconds to the period of the "good" clock in order to get the period of the slower clock. The reverse applies for the clock that gains time. Your expressions have the signs reversed.

 

[Chestermiller]

I am not so sure about this. Imagine a clock that displays time in seconds over a 24-hour period, from 0 to 86,399. A clock that loses 12 seconds over a 24-hour period will display 86,387 when the accurate clock displays 86,399. For that to happen, the period of the "slow" pendulum that regulates its timing must be longer, not shorter, than that of the "good" clock. In other words one must add $\frac{12}{86,400}$ seconds to the period of the "good" clock in order to get the period of the slower clock. The reverse applies for the clock that gains time. Your expressions have the signs reversed.

Oh. I made an algebra error, which I have now gone back and corrected. Thanks.

 

[Chestermiller]

The period of the pendulum is given by $$P=2\pi\sqrt{\frac{L}{g}}$$From this, if follows that $$d\ln{P}=\frac{1}{2}d\ln{L}=\frac{1}{2}\alpha \Delta T$$or $$\ln{\left[\frac{(1+0.000139)}{(1-0.0000463)}\right]}=\frac{1}{2}\alpha(20)$$

 

[haruspex]

actually for the jee answer to be correct this should be the interpretation of gaining time. which is slight different for the interpretation of losing time. losing seconds means the actual seconds(not slow pendulum seconds) that are needed for the slower pendulum to strike the 24th hour right after the accurate pendulum strikes it. And now for gaining seconds this is the meaning for answer to be correct, The number of seconds(real seconds again) it takes for the accurate pendulum to strike the 24th hour right after the faster pendulum struck the 24th hour.

To be clearer, let's express it in ratios. If N is the number of seconds in a day and x is the number gained in a day (could be negative), is the clock rate $\frac{N+x}N$ times correct speed or $\frac N{N-x}$ times correct speed? Or compromise at $\sqrt\frac{N+x}{N-x}$?
Is JEE being consistent?