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Reflection and Transmission of Waves On a String

Reflection and transmission of waves On a string :- When a wave traveling along a string encounters a boundary or interface, such as the end of the string or a point where the string is attached to a different material, partial reflection and partial transmission occurs. Since the tensions are the same, the relative magnitudes of the wave velocities depends on the mass densities (μ) of the ropes.

For instance, suppose a light string is attached to a heavier string as shown in above figure (a). When a pulse travelling on the light string reaches the knot, some part of it is reflected and inverted and some part of it is transmitted to the heavier string. As expected, the reflected pulse has a smaller amplitude than the incident pulse, because a part of the incident energy is transferred to the transmitted pulse in the heavier string. The inversion in the reflected wave is because of the reflection from a rigid boundary.

In below figure (b), a pulse travelling along a heavy string approaches the boundary of a lighter string. The light string offers little resistance and now behaves like a free boundary. Consequently, the reflected pulse is not inverted.

In either case, the relative height of the reflected and transmitted pulses depend on the relative densities of the two string.

Boundary Conditions

(Reflection and Transmission of Waves On a String)

Let us study the division of the incident wave into reflected and transmitted components mathematically and derive the expressions for reflected wave amplitude, transmitted wave amplitude, reflection coefficient (R) and transmission coefficient (T). To derive these expressions let us consider two strings of liner mass densities μ₁ and μ₂ joined at x = 0 as shown in figure below.

Both the incident and reflected pulses travel on the left string with a speed $\displaystyle {{v}_{1}}\text{=}\sqrt{\frac{T}{{{\mu }_{1}}}}$ while the transmitted pulse travels to along the right string with a speed $\displaystyle {{v}_{2}}\text{=}\sqrt{\frac{T}{{{\mu }_{2}}}}$ .

Let the incident, reflected and transmitted waves are represented by y_in , y_r and y_t respectively. The equations for these waves can be written as :

$\displaystyle {{y}_{in}}={{A}_{in}}\sin (\omega t-{{k}_{1}}x)$ …..(1)

$\displaystyle {{y}_{r}}={{A}_{r}}\sin (\omega t+{{k}_{1}}x)$ …..(2)

$\displaystyle {{y}_{t}}={{A}_{t}}\sin (\omega t-{{k}_{2}}x)$ …..(3)

Here k₁ and k₂ are the wave numbers of the different media and we also know that

$\displaystyle {{v}_{1}}=\frac{\omega }{{{k}_{1}}}$ …..(4)

$\displaystyle {{v}_{2}}=\frac{\omega }{{{k}_{2}}}$ …..(5)

Now there are two independent conditions which the pulses must satisfy at the boundary, these are :

(1) The net vertical displacement on both sides of the boundary must be equal at all times, i.e.,

$\displaystyle {{y}_{in}}(0,t)+{{y}_{r}}(0,t)={{y}_{t}}(0,t)$ …..(6)

(2) The two strings must exert equal and opposite forces on each other at the junction, i.e., $\displaystyle \overrightarrow{{{F}_{1}}}=-\overrightarrow{{{F}_{2}}}$ as shown in figure below :

This condition only holds if the slops of the two strings are the same at the junction point, i.e.,

$\displaystyle {{\left\{ \frac{\partial }{\partial x}[{{y}_{in}}(x,t)+{{y}_{r}}(x,t)] \right\}}_{x=0}}={{\left\{ \frac{\partial }{\partial x}[{{y}_{t}}(x,t)] \right\}}_{x=0}}$ …..(7)

Applying the first boundary condition :-

$\displaystyle L.H.S.={{y}_{in}}(0,t)+{{y}_{r}}(0,t)={{A}_{in}}\sin (\omega t-{{k}_{1}}0)+{{A}_{r}}\sin (\omega t+{{k}_{1}}0)$

$\displaystyle \Rightarrow L.H.S.={{A}_{in}}(sin\omega t)+{{A}_{r}}(sin\omega t)$

$\displaystyle \therefore L.H.S.=\left( {{A}_{in}}+{{A}_{r}} \right)sin\omega t$ …..(8)

$\displaystyle R.H.S.={{y}_{t}}(0,t)={{A}_{t}}\sin (\omega t-{{k}_{2}}0)$

$\displaystyle \therefore R.H.S.={{A}_{t}}sin\omega t$ …..(9)

From equations (8) and (9),

$\displaystyle {{A}_{t}}={{A}_{in}}+{{A}_{r}}$ …..(10)

Applying the secondary boundary condition :-

$\displaystyle {{\left\{ \frac{\partial }{\partial x}[{{y}_{in}}(x,t)+{{y}_{r}}(x,t)] \right\}}_{x=0}}={{\left\{ \frac{\partial }{\partial x}[{{y}_{t}}(x,t)] \right\}}_{x=0}}$

$\displaystyle {{\left\{ \frac{\partial }{\partial x}[{{A}_{in}}sin(\omega t-{{k}_{1}}x)+{{A}_{r}}sin(\omega t+{{k}_{1}}x)] \right\}}_{x=0}}={{\left\{ \frac{\partial }{\partial x}[{{A}_{t}}sin(\omega t-{{k}_{2}}x)] \right\}}_{x=0}}$

$\displaystyle {{\left[ -{{k}_{1}}{{A}_{in}}cos(\omega t-{{k}_{1}}x)+{{k}_{1}}{{A}_{r}}cos(\omega t+{{k}_{1}}x) \right]}_{_{x=0}}}={{\left[ -{{k}_{2}}{{A}_{t}}cos(\omega t-{{k}_{2}}x) \right]}_{_{_{x=0}}}}$

$\displaystyle \Rightarrow \left[ -{{k}_{1}}{{A}_{in}}cos\omega t+{{k}_{1}}{{A}_{r}}cos\omega t \right]=\left[ -{{k}_{2}}{{A}_{t}}cos\omega t \right]$

$\displaystyle -{{k}_{1}}{{A}_{in}}+{{k}_{1}}{{A}_{r}}=-{{k}_{2}}{{A}_{t}}$ …..(11)

Using the value of A_t from equation (10) in equation (11), we get

$\displaystyle -{{k}_{1}}{{A}_{in}}+{{k}_{1}}{{A}_{r}}=-{{k}_{2}}\left( {{A}_{in}}+{{A}_{r}} \right)$

$\displaystyle -{{k}_{1}}{{A}_{in}}+{{k}_{1}}{{A}_{r}}=-{{k}_{2}}{{A}_{in}}-{{k}_{2}}{{A}_{r}}$

$\displaystyle \Rightarrow {{k}_{1}}{{A}_{r}}+{{k}_{2}}{{A}_{r}}={{k}_{1}}{{A}_{in}}-{{k}_{2}}{{A}_{in}}$

$\displaystyle \left( {{k}_{1}}+{{k}_{2}} \right){{A}_{r}}=\left( {{k}_{1}}-{{k}_{2}} \right){{A}_{in}}$

Hence reflected wave amplitude :

$\displaystyle {{A}_{r}}=\frac{\left( {{k}_{1}}-{{k}_{2}} \right)}{\left( {{k}_{1}}+{{k}_{2}} \right)}{{A}_{in}}=\frac{\left( {{\upsilon }_{2}}-{{\upsilon }_{1}} \right)}{\left( {{\upsilon }_{2}}+{{\upsilon }_{1}} \right)}{{A}_{in}}$ …..(12)

The reflection coefficient (𝑅) i.e., the ratio of the amplitude of the reflected wave (𝐴_𝑟) to the amplitude of the incident wave (𝐴_𝑖) is given by :

$\displaystyle R=\frac{{{A}_{r}}}{{{A}_{in}}}=\frac{\left( {{k}_{1}}-{{k}_{2}} \right)}{\left( {{k}_{1}}+{{k}_{2}} \right)}$ …..(13)

Substituting the value of A_r from equation (12) in equation (10), we get

$\displaystyle {{A}_{t}}={{A}_{in}}+{{A}_{r}}={{A}_{in}}+\frac{\left( {{k}_{1}}-{{k}_{2}} \right)}{\left( {{k}_{1}}+{{k}_{2}} \right)}{{A}_{in}}$

$\displaystyle {{A}_{t}}=\left[ 1+\frac{\left( {{k}_{1}}-{{k}_{2}} \right)}{\left( {{k}_{1}}+{{k}_{2}} \right)} \right]{{A}_{in}}$

$\displaystyle {{A}_{t}}=\left[ \frac{{{k}_{1}}+{{k}_{2}}+{{k}_{1}}-{{k}_{2}}}{\left( {{k}_{1}}+{{k}_{2}} \right)} \right]{{A}_{in}}$

Hence transmitted wave amplitude :

$\displaystyle {{A}_{t}}=\left( \frac{2{{k}_{1}}}{{{k}_{1}}+{{k}_{2}}} \right){{A}_{in}}=\left( \frac{2{{\upsilon }_{2}}}{{{\upsilon }_{2}}+{{\upsilon }_{1}}} \right){{A}_{in}}$ …..(14)

The transmission coefficient (T) i.e., the ratio of the amplitude of the transmitted wave (𝐴_t) to the amplitude of the incident wave (𝐴_𝑖) is given by :

$\displaystyle T=\frac{{{A}_{t}}}{{{A}_{in}}}=\left( \frac{2{{k}_{1}}}{{{k}_{1}}+{{k}_{2}}} \right)$ …..(15)

Reflection and Transmission of Waves On a String

Boundary Conditions

(Reflection and Transmission of Waves On a String)

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