Question

A harmonic wave $y_i=0.002cos(2\pi x-50\pi t)$ travels along a string towards a boundary at x = 0, with a second string. The wave speed on the second string is $50m{s}^{-1}$. The expression for transmitted wave is (Assume SI units)

• A. $y_t=(2.67\times {10}^{-4})cos(2\pi x-50\pi t)$
• B. $y_t=(2.67\times {10}^{-4})cos(\pi x-50\pi t)$
• C. $y_t=(2.67\times {10}^{-3})cos(2\pi x-50\pi t)$
• D. $y_t=(2.67\times {10}^{-3})cos(\pi x-50\pi t)$

Answer 1

The correct option is D $y_t=(2.67\times {10}^{-3})cos(\pi x-50\pi t)$

$v_1=\frac{{\omega }_1}{k_1}=\frac{50\pi }{2\pi }=25m{s}^{-1}$ and $v_2=50m{s}^{-1}\left(given\right)$
$A_t=\left(\frac{2v_2}{v_1+v_2}\right)A_i=\frac{2\times 50}{25+50}\left(0.002\right)=2.67\times {10}^{-3}$
${\omega }_1={\omega }_2(\because frequencydependsonsource)$
$\therefore k_2=\frac{{\omega }_2}{v_2}=\frac{{\omega }_1}{v_2}=\frac{50\pi }{50}=\pi$
$\therefore y_t=(2.67\times {10}^{-3}cos(\pi x-50\pi t)$