Question

The equation of the transverse wave travelling in a rope is given by $y=5\sin (4t-0.02x)$ where $y$ and $x$ are in meters and $t$ is in seconds. Calculate the intensity of the wave if the density of rope material is $1250kg/m^3$

• A. $2000kJ{m}^{-2}{s}^{-1}$
• B. $10kJ{m}^{-2}{s}^{-1}$
• C. $40kJ{m}^{-2}{s}^{-1}$
• D. $3kJ{m}^{-2}{s}^{-1}$

Answer 1

The correct option is A $2000kJ{m}^{-2}{s}^{-1}$

Comparing the given equation with the standard equation of wave motion
$y=a\sin (2\pi nt-\frac{2\pi x}{\lambda })$
We get, amplitude $a=5cm$ and $2\pi n=4$
$⟹n=\frac{4}{2\pi }=0.637Hz$
$\lambda =\frac{2\pi }{0.02}=3.14m$
Density $\rho =1250kg/m^3$
Velocity of wave $\nu =n\lambda =\frac{4}{2\pi }\times \frac{2\pi }{0.02}=200m/sec$
Intensity of wave$=\frac{1}{2}\rho \nu (2\pi n{)}^2$
$=\frac{1}{2}\times 1250\times 16\times 200$
$=2000000J{m}^{-2}{s}^{-1}$
$=2000kJ{m}^{-2}{s}^{-1}$