Question

A standing wave $\zeta =\left(5mm\right)\sin \pi x.\cos \left(200t\right)$ is maintained in a homogeneous rod with cross-sectional area $0.04m^2$ and density $1000kg/m^3.$ The total mechanical-energy confined between the sections corresponding to the adjacent displacement nodes is

• A. $10J$
• B. $20J$
• C. $40J$
• D. None of these

Answer 1

The correct option is A $10J$

$Totalmechanicalenergyofvibration=\int \frac{1}{2}dm{v}^{2}dm=dx,={10}^{3}kg/m^3\times \left(0.04\right)m^2=40kg/mv\left(x\right)=ds/dt=5\left(200\right)\sin \pi x\left(sin\right(200t\left)\right)Totalmechanicalenergy=\frac{1}{2}\int v^2\left(x\right)dx=10J.$