Question

A harmonic wave is travelling on string $1$. At the junction with string $2$, it is partly reflected and partly transmitted. The linear mass density of string $2$ is $9$ times that of string $1$, and that the boundary between the two strings is at $x=0$. If the expression for the incident wave is
$y_i=\left(3\text{cm}\right)\cos \left[\right(3.14{\text{cm}}^{-1})x-(314{\text{rad s}}^{-1}\left)t\right]$. What is the expression for the transmitted waves?

• A. $y_t=\left(3\text{cm}\right)\cos \left[\right(3.14{\text{cm}}^{-1})x-(314{\text{rad s}}^{-1}\left)t\right)]$
• B. $y_t=\left(1.5\text{cm}\right)\cos \left[\right(3{\text{cm}}^{-1})x-(9.42{\text{rad s}}^{-1}\left)t\right]$
• C. $y_t=\left(3\text{cm}\right)\cos \left[\right(9.42{\text{cm}}^{-1})x-(3.14{\text{rad s}}^{-1}\left)t\right]$
• D. $y_t=\left(1.5\text{cm}\right)\cos \left[\right(9.42{\text{cm}}^{-1})x-(314{\text{rad s}}^{-1}\left)t\right]$

Answer 1

The correct option is D $y_t=\left(1.5\text{cm}\right)\cos \left[\right(9.42{\text{cm}}^{-1})x-(314{\text{rad s}}^{-1}\left)t\right]$

Given equation of incident wave
$y_i=3\text{cm}\cos \left[\right(3.14{\text{cm}}^{-1})x-(314{\text{rad s}}^{-1}\left)t\right]$
So, $A_i=3\text{cm},k_1=3.14{\text{cm}}^{-1},{\omega }_1=314{\text{rad s}}^{-1}$
and
Linear mass density $\left(\mu \right)$ and tension $T$ on the string
${\mu }_2=9{\mu }_1$ and $T_1=T_2=T$
We know,
Wave speed $\left(v\right)=\sqrt{\frac{T}{\mu }}$
So, $v_1=\sqrt{\frac{T}{{\mu }_1}}\&v_2=\sqrt{\frac{T}{{\mu }_2}}=\sqrt{\frac{T}{9{\mu }_1}}=\frac{v_1}{3}$

As we know that, frequency of wave does not change
i.e ${\omega }_1={\omega }_2=\omega =314{\text{rad s}}^{-1}$
Wave number $\left(k\right)=\frac{\omega }{v}$
$\Rightarrow k_2=\frac{{\omega }_2}{v_2}=\frac{{\omega }_1}{\frac{v_1}{3}}[\because v_2=\frac{v_1}{3}]$
$\Rightarrow k_2=3\frac{{\omega }_1}{v_1}=3k_1=3\times 3.14=9.42{\text{cm}}^{-1}$
Now, amplitude $\left(A_t\right)$ of transmitted wave
$A_t=\frac{2v_2}{v_1+v_2}{A}_i=\frac{2v_2}{3v_2+v_2}\left(3\text{cm}\right)=1.5\text{cm}$

Hence, equation of transmitted wave can be written as
$y_t=\left(1.5\text{cm}\right)\cos \left[\right(9.42{\text{cm}}^{-1})x-(314{\text{rad s}}^{-1}\left)t\right]$