Question

Two strings A and B with $\mu = 2\dfrac{{kg}}{m}$ and $\mu = 8\dfrac{{kg}}{m}$ respectively are joined in series and kept on a horizontal table with both the ends fixed. The tension in the string is 200 N. If a pulse of an amplitude 1 cm travels in A towards the junction, then find the amplitude of the reflected and transmitted pulse.
A) $A = - \dfrac{1}{3}cm,A = \dfrac{2}{3}cm$
B) $A = - \dfrac{1}{2}cm,A = \dfrac{2}{3}cm$
C) $A = - \dfrac{1}{3}cm,A = \dfrac{1}{3}cm$
D) $A = - \dfrac{2}{3}cm,A = \dfrac{1}{3}cm$

Answer 1

Hint: To solve this problem, firstly, we have to find the velocity of the wave in each string by using the formula:
$V = \sqrt {\dfrac{T}{\mu }} $
where T is the tension and μ is the mass per unit length of a string.
We have to use this velocity to calculate the wave constant, in order to evaluate the amplitudes of the transmitted and the reflected waves.

Complete step by step answer:
Let \[{V_A}\] and \[{V_B}\] are the velocities of the wave on string A and string B respectively similarly \[{K_A}\]and \[{K_B}\]be their wave constants.
Given:
The mass per unit length of the string A: ${\mu _A} = 2\dfrac{{kg}}{m}$
The mass per unit length of the string B: ${\mu _B} = 8\dfrac{{kg}}{m}$
Amplitude (A) = 1 cm and \[T = 200{\text{ N}}\]
Step I
Calculate the velocities of string waves on each string.
Formula used: $V = \sqrt {\dfrac{T}{\mu }} $
The velocity of the wave on string A
\[{V_A} = \sqrt {\dfrac{T}{{{\mu _A}}}} \]……………..(i)
Substitute the given values in eqn (i), we get
\[{V_A} = \sqrt {\dfrac{{200}}{2}} \]
\[ \Rightarrow {V_A} = 10{\text{ }}m/s\]
Similarly, the Velocity of the wave on string B will be:-
\[{V_B} = \sqrt {\dfrac{{200}}{8}} {\text{ }}m/s\]
\[ \Rightarrow {V_B} = 5{\text{ }}m/s\]

Step II
Let the wave frequency be\[\omega \].
Calculate the wave constants \[{K_A}\]and \[{K_B}\] for a wave on string A and String B respectively.
Formula used: \[k = \dfrac{\omega }{V}\]
For wave on string A
\[{K_A} = \dfrac{\omega }{{{V_A}}}\]…… ……………..(ii)
Substituting the given values in eqn (ii), we get
\[ \Rightarrow {K_A} = 0.1\omega \]…………….. (iii)
Similarly, for wave B
\[ \Rightarrow {K_B} = \dfrac{\omega }{5}\]
\[ \Rightarrow {K_B} = 0.2\omega \]……………… (iv)

Step III
Calculate the amplitude of the reflected and transmitted wave.
The amplitude of reflected wave-
\[{A_R} = \left( {\dfrac{{{K_A} - {K_B}}}{{{K_A} + {K_B}}}} \right)A\]……………..(v)
Using eqn (iii), eqn (iv), and the given value of A in eqn (v), we get
\[{A_R} = \left( {\dfrac{{0.1 - 0.2}}{{0.1 + 0.2}}} \right) \times 0.1{\text{ m}}\]
\[ \Rightarrow {A_R} = - \dfrac{1}{3}\]…………….. (vi)
Now,
The amplitude of Transmitted wave-
\[{A_T} = A - \left| {{A_R}} \right|\]………….. (vii)
Using eqn (vi) in eqn (vii)
\[ \Rightarrow {A_T} = 1 - \dfrac{1}{3}\]
\[ \Rightarrow {A_T} = \dfrac{2}{3}\]
Therefore, the amplitude of reflected and the transmitted wave are \[ - \dfrac{1}{3}\] and \[\dfrac{2}{3}\].

Hence, Option (A) is the correct answer.

Note: To tackle these kinds of questions the key is to practice a lot of formula-based questions on this topic and make their own short-notes and formula while deriving the important results. One should not confuse the formula for wave constant for sound is different from the wave constant for non-mechanical waves. The formula will remain the same only the values for the quantities involved will differ.