0
$\begingroup$

The length of a pendulum is measured using a meter scale which has 2000 divisions. If the measured value L is 50cm, the accuracy in the determination of g is 1.1% and the time taken for 100 oscillations is 100 seconds, what should be the resolution of the clock (in milliseconds).

I have a confusion about calculation of fractional error in time period.Shouldn't the fractional errors in length and gravity be added and divided by two to find fractional error of time period?

My solution is :

percent error in length = (100cm/2000 i.e least count)/50cm x 100 =0.1.

percent error in g = 1.1 percent(given)

percent error in T = 1/2 (0.1 +1.1)=0.6 (this is obtained by taking logarithms and differentiation)

fractional error in T = resolution/total time =0.6/100 so resolution is 0.6 seconds or 600 milliseconds .

However the answer given is 5ms.

Can someone give me some clarity about fractional errors in time period ? Is my procedure rational? Thanks in advance.

$\endgroup$
4
  • $\begingroup$ $\dfrac{\Delta T}{T}=\dfrac{1}{2}\cdot \dfrac{\Delta L}{L}+\dfrac{1}{2}\cdot \dfrac{\Delta g}{g}$ $\endgroup$
    – user279106
    Aug 5, 2021 at 6:26
  • $\begingroup$ Yes,I have used that equation , so is my answer right? $\endgroup$ Aug 5, 2021 at 6:49
  • $\begingroup$ Is this a right way to calculate resolution of watch? $\endgroup$ Aug 9, 2021 at 2:58
  • $\begingroup$ You may think the error in timing is very small but it is usual to time a number (20?) oscillations with say an error of 100 ms so the error in one oscillation would be 100/20 = 5 ms. $\endgroup$
    – Farcher
    Feb 11, 2022 at 23:03

1 Answer 1

2
+50
$\begingroup$

the time period T is a function of the length L and the gravitation g :

with: $$ T\propto\sqrt{\frac Lg}\\ dT=\frac{\partial T}{\partial L}\,dL+\frac{\partial T}{\partial g}\,dg\quad\Rightarrow\\ \frac{dT}{T}=\frac 12\left(\frac {dL}{L}-\frac{dg}{g}\right)$$

with your data $~\frac {dL}{L}=0.1~,\frac {dg}{g}=0.01~$ you obtain $~\frac {dT}{T}=0.045~=4.5\%$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.