Question

The speed of a longitudinal wave in a metallic bar is \(400\,\text{m/s}\). If the density and Young’s modulus of the bar material increase by \(0.5%\) and \(1%\) respectively, then the speed of the wave is changed approximately to _______ m/s.

• A. \(399\)
• B. \(398\)
• C. \(402\)
• D. \(401\)

Hint:

For quantities under square root, remember that relative error is halved.

Answer 1

The Correct Option is D

Concept: The speed of a longitudinal wave in a rod is: \[ v=\sqrt{\frac{Y}{\rho}} \] where \(Y\) is Young’s modulus and \(\rho\) is density.
Step 1: Relative error formula Taking logarithmic differentiation: \[ \frac{\Delta v}{v} =\frac12\left(\frac{\Delta Y}{Y}-\frac{\Delta \rho}{\rho}\right) \]
Step 2: Substitute given percentage changes \[ \frac{\Delta Y}{Y}=1%=0.01,\quad \frac{\Delta \rho}{\rho}=0.5%=0.005 \] \[ \frac{\Delta v}{v} =\frac12(0.01-0.005)=0.0025 \]
Step 3: Calculate new speed \[ \Delta v=400\times0.0025=1\,\text{m/s} \] \[ v_{\text{new}}=400+1=401\,\text{m/s} \] Final Answer: \[ \boxed{401\ \text{m/s}} \]