Question

The harmonic wave $y_i=(2.0\times {10}^{-3})cos\pi (2.0x-50t)$ travels along a string toward a boundary at $x=0$ with a second string. The wave speed on the second string is $50m/s$. What are the expressions for reflected and transmitted waves. Assume SI units

• A. $6.67\times {10}^{-4}cos\pi (2.0x+50t),2.67\times {10}^{-3}cos\pi (x-50t)$
• B. $6.67\times {10}^{-4}cos\pi (2.0x-50t),2.67\times {10}^{-3}cos\pi (x-50t)$
• C. $6.67\times {10}^{-4}cos\pi (2.0x+50t+\pi ),2.67\times {10}^{-3}cos\pi (x-50t)$
• D. $6.67\times {10}^{-4}cos\pi (2.0x-50t+\pi ),2.67\times {10}^{-3}cos\pi (x-50t)$

Answer 1

The correct option is A

$6.67\times {10}^{-4}cos\pi (2.0x+50t),2.67\times {10}^{-3}cos\pi (x-50t)$

In such situation a part of the wave gets reflected while some gets transmitted. The amplitude of reflected wave is given by

$A_r=\left(\frac{v_2-v_1}{v_1+v_2}\right)A_i$ where $v_2$ and $v_1$ are velocities in second and first medium respectively.

Here $v_1=\frac{\omega }{k_1}=\frac{50\pi }{2\pi }=25m{s}^{-1}$ from the given equation.

$y_i=(2.0\times {10}^{-3})cos\left[\pi \right(2.0x-50t\left)\right]$

$v_2=50m{s}^{-1}$

$A_r=\left(\frac{50-25}{50+25}\right)(2\times {10}^{-3}$

= $6.67\times {10}^{-4}m$

So the equation of reflected wave would be

$y_r=6.67\times {10}^{-4}cos\left[\pi \right(2.0x+50t\left)\right]$

as the wave is going from higher density medium to lower density medium, so there is no phase change

In the second medium,

$k_2=\frac{\omega }{v_2}$[frequency does not change]

= $\frac{50\pi }{50}=\pi m^{-1}$

Also. $A_t=\left(\frac{2v_2}{v_1+v_2}\right)A_i$

= $\left(\frac{2\times 50}{25+50}\right)\times 2\times {10}^{-3}$

= $2.67\times {10}^{-3}m$

So the wave equation is $y_r=2.67\times {10}^{-3}cos\left[\pi \right(x-50t\left)\right]$