Physics And Measurement Question 302
Question: The frequency (f) of a wire oscillating with a length $ \ell $ , in p loops, under a tension T is given by $ f=\frac{P}{2\ell }\sqrt{\frac{T}{\mu }} $ where $ \mu = $ linear density of the wire. If the error made in determining length, tension and linear density be 1%, -2% and 4%, then find the percentage error in the calculated frequency
Options:
A) - 4%
B) - 2%
C) - 1%
D) - 5%
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Given $ f=\frac{\rho }{2\ell }\sqrt{\frac{T}{\mu }} $
Taking log of both sides $ \log f=\log ( \frac{p}{2} )-\log \ell +\frac{1}{2}\log T-\frac{1}{2}\log \mu $
Differentiating partially on both sides,
$ \frac{df}{f}=0-\frac{d\ell }{\ell }+\frac{1}{2}\frac{dT}{T}-\frac{1}{2}\frac{d\mu }{\mu } $
or $ \frac{df}{f}\times 100=( -\frac{d\ell }{\ell }\times 100 )+( \frac{1}{2}\frac{dT}{T}\times 100 )-( \frac{1}{2}\frac{d\mu }{\mu }\times 100 ) $
$ =(-1)+\frac{1}{2}(-2)-\frac{1}{2}(4)=-1-1-2=-4% $
BETA
