Question 46: Units , Dimensions and Error Analysis

Question

The speed of a longitudinal wave in a metallic bar is 400 m/s. If the density and Young's modulus of the bar material are increased by 0.5% and 1%, respectively then the speed of the wave is changed approximately to ______ m/s.

Accepted Answer

The formula for the speed of a longitudinal wave ( v ) in a solid metallic bar is given as
$ \mathrm{v}=\sqrt{\frac{\mathrm{Y}}{ho}} $
Where, $\mathrm{Y}=$ Young's modulus and its $ho$ is its density
Taking the natural $\log$ on both sides of $\mathrm{v}=\left(\frac{\mathrm{Y}}{ho}ight)^{1 / 2}$ :
$ \ln (\mathrm{v})=\frac{1}{2} \ln (\mathrm{Y})-\frac{1}{2} \ln (ho) $
Differentiating both sides gives the fractional change:
$ \frac{\Delta v}{v}=\frac{1}{2} \frac{\Delta Y}{Y}-\frac{1}{2} \frac{\Delta ho}{ho} $
Initial speed $=\mathrm{v}=400 \mathrm{~m} / \mathrm{s}$
Increase in Young's modulus $=\frac{\Delta \mathrm{Y}}{\mathrm{Y}} 	imes 100=+1 \%$
Increase in density $\frac{\Delta ho}{ho} 	imes 100=+0.5 \%$
So, the percentage change in speed,
$ \% 	ext { change in } \mathrm{v}=\frac{1}{2}(1 \%)-\frac{1}{2}(0.5 \%)=0.5 \%-0.25 \%=0.25 \% $
Since the result is positive, the speed increases by $0.25 \%$.
So, new speed $\left(\mathrm{v}^{\prime}ight)=$ Initial Speed + Change in Speed
$ \begin{aligned} \mathrm{v}^{\prime} & =\mathrm{v}+(0.25 \% 	ext { of } \mathrm{v}) \ \Rightarrow \mathrm{v}^{\prime} & =400+\left(\frac{0.25}{100} 	imes 400ight) \ \Rightarrow \mathrm{v}^{\prime} & =400+1=401 \mathrm{~m} / \mathrm{s} \end{aligned} $
Hence, the correct option is (C).

Answer

398

Answer

402

Answer

401

Answer

399

Units , Dimensions and Error Analysis

The speed of a longitudinal wave in a metallic bar is 400 m/s. If the density and Young's modulus of the bar material are increased by 0.5% and 1%, respectively then the speed of the wave is changed approximately to ______ m/s.

Choose the correct answer: