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.

• A. $-4 \%$
• B. $-2 \%$
• C. $-1 \%$
• D. $-5 \%$

Answer 1

Answer: (A)

Given $f=\frac{p}{2 \ell} \sqrt{\frac{T}{\mu}}$. Taking log of both sides
$\log f=\log \left(\frac{ p }{2}\right)-\log \ell+\frac{1}{2} \log T -\frac{1}{2} \log \mu$
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=\left(-\frac{ d \ell}{\ell} \times 100\right)+\left(\frac{1}{2} \frac{ dT }{ T } \times 100\right)-\left(\frac{1}{2} \frac{ d \mu}{\mu} \times 100\right) $
$=(-1)+\frac{1}{2}(-2)-\frac{1}{2}(4)$
$=-1-1-2=-4 \%$