Superposition of Waves
6.0 Questions
6.0 Questions
Question 1. Two harmonic waves are represented in SI units by,$${y_1}(x,t) = 0.2\sin (x - 3.0t)$$$${y_2}(x,t) = 0.2\sin (x - 3.0t + \phi )$$(a) Write the expression for the sum $y = {y_1} + {y_2}$ for $\phi = \frac{\pi }{2}$ rad.
(b) Suppose the phase difference $\phi $ between the waves is unknown and the amplitude of their sumis .032$m$, what is $\phi $?
Solution:
(a) $$y = {y_1} + {y_2}$$$$ = 0.2\sin (x - 3.0t) + 0.2\sin \left( {x - 3.0t + \frac{\pi }{2}} \right)$$$$ = A\sin (x - 3.0t + \theta )$$Here, $$A = \sqrt {{{(0.2)}^2} + {{(0.2)}^2}} = 0.28m$$and $$\phi = \frac{\pi }{4}$$$$\therefore y = 0.28\left( {x - 3.0t + \frac{\pi }{4}} \right)$$
(b) Since the amplitude of the resulting wave is$0.32m$ and $A=.02m$, $$0.32 = \sqrt {{{(0.2)}^2} + {{(0.2)}^2} + (2)(0.2)(0.2)\cos \phi } $$Solving,$$\phi = \pm 1.29rad$$
Question 2.
Two waves of equal frequencies have their amplitudes in the ratio of $3:5$. They are superimposed on each other. Calculate the ratio of maximum and minimum intensities of the resultant wave?
Solution: Given,$$\frac{{{A_1}}}{{{A_2}}} = \frac{3}{5}$$$$\therefore \sqrt {\frac{{{I_1}}}{{{I_2}}}} = \frac{3}{5}$$
Maximum intensity is obtained, where$${I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}$$
Minimum intensity is found, where $$\cos \phi = - 1$$$${I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}$$
Hence, $$\frac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\frac{{\sqrt {{I_1}} + \sqrt {{I_2}} }}{{\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2} = \left( {\frac{{\sqrt {\frac{{{I_1}}}{{{I_2}}}} + 1}}{{\sqrt {\frac{{{I_1}}}{{{I_2}}}} - 1}}} \right)$$$$ = {\left( {\frac{{\frac{3}{5} + 1}}{{\frac{3}{5} - 1}}} \right)^2} = \frac{{64}}{4} = 16$$
Question 3. Two strings 1 and 2 are taut between two fixed supports such that the tension in both strings is same. Mass per unit length of 2 is more than that of 1. Explain which string is denser for a transverse traveling wave.
Soution:-
Speed of a transverse wave on a string$$v = \sqrt {\frac{T}{\mu }} $$or$$v \propto \frac{1}{{\sqrt \mu }}$$Now,$${\mu _2} > {\mu _1}$$$$\therefore {v_2} < {v_1}$$
i.e., medium 2 is denser and medium 1 is rarer.
Question 4.
A triangular pulse moving at $2cm/s$ on a rope approaches an end at which it is free to slide on a vertical pole.
(a) Draw the pulse at $\frac{1}{2}$ interval until it is completely reflected.
(b) What is the particle speed on the trailing edge at the instant depicted?
Solution:
(a) Reflection of a pulse from a free boundary is really the superposition of two identical waves traveling in opposite direction.
In every $\frac{1}{2}s$, each pulse ( one real moving towards the right and one imaginary moving towards left ) travels a distance of $1 cm$, as the wave speed is $2 cm/s$.
(b) Particle speed, $${v_p} = \left| { - v(slope)} \right|$$ Here, $v$= wave speed= $2 cm/s$ and slope = $\frac{1}{2}$
Particle speed = $1 cm/s$
Question 5. The displacement of a standing wave on a string is given by $$y(x,t) = 0.4\sin (0.5x)\cos (30t)$$
where $x$ and $y$ are in centimeters.
(a) Find the frequency, amplitude and wave speed of the component waves.
(b) What is the particle velocity at $x=2.4 cm$ at $t=0.8s$?
Solution:
(a) The given wave can be written as the sum of two component waves as$$y(x,t) = 0.2\sin (0.5x - 30t) + 0.2\sin (0.5x + 30t)$$
That is the two component waves are $${y_1}(x,t) = 0.2\sin (0.5x - 30t)$$ travelling in positive $x$-direction and $${y_2}(x,t) = 0.2\sin (0.5x + 30t)$$ travelling in negative $x$-direction
Question 6. The vibrations of a string of length $60cm$ fixed at both ends are represented by the equation $y = 4\sin \left( {\frac{{\pi x}}{{15}}} \right)\cos \left( {96\pi t} \right)$ where $x$ and $y$ are in $cm$ and$t$ is in seconds.
(i) What is the maximum displacement of a point at $x=5cm$?
(ii) Where are the nodes located along the string?
(iii) What is the velocity of the particle at $x=7.5cm$ and at $t=0.25s$?
(iv) Write down the equations of the component waves whose superposition gives the above wave.
Solution:
$$y = 4\sin \left( {\frac{{\pi x}}{{15}}} \right)\cos \left( {96\pi t} \right)$$can be broken up into
$$y = 2\left[ {\sin \left( {\frac{{\pi x}}{{15}} + 96\pi t} \right) + \sin \left( {\frac{{\pi x}}{{15}} - 96\pi t} \right)} \right]$$
Thus, the waves are of the same amplitude and frequency but travelling in opposite directions which, therefore, superimpose to give a standing wave.
(i) At $x=5cm$, the standing wave equation gives $$y = 4\sin \left( {\frac{{5\pi }}{{15}}} \right)\cos \left( {96\pi t} \right)$$$$ = 4\sin \frac{\pi }{3}\cos (96\pi t) = 4 \times \sqrt {\frac{3}{2}} \cos (96\pi t)$$
$$\therefore {\text{Maximum displacement = 2}}\sqrt 3 cm$$
(ii) The nodes are the points permanently at rest. Thus, they are those points for which $\sin \left( {\frac{{\pi x}}{{15}}} \right) = 0$.
$$i.e.\frac{{\pi x}}{{15}} = n\pi ,n = 0,1,2,3....$$$$ \Rightarrow x = 0,15,30,45....$$
(iii) The particle velocity is equal to $$\left( {\frac{{\delta y}}{{\delta t}}} \right) = 4\sin \left( {\frac{{\pi x}}{{15}}} \right)\left( {96\pi } \right)\left( { - \sin 96\pi t} \right)$$$$ = - 384\pi \sin \left( {\frac{{\pi x}}{{15}}} \right)\sin (96\pi t)$$
At $x=7.5$ and $t=0.25$,$$\begin{equation} \begin{aligned} \left( {\frac{{\delta y}}{{\delta t}}} \right) = - 384\pi \sin \left( {\frac{{\pi x}}{{15}}} \right)\sin (96\pi t) \\ \left( {\frac{{\delta y}}{{\delta t}}} \right) = - 384\pi \sin \left( {\frac{\pi }{2}} \right)\sin (24\pi ) \\ \left( {\frac{{\delta y}}{{\delta t}}} \right) = 0 \\\end{aligned} \end{equation} $$
(iv) The equations of the component waves are$${y_1} = \sin \left( {\frac{{\pi x}}{{15}} + 96\pi t} \right)$$and$${y_2} = \sin \left( {\frac{{\pi x}}{{15}} - 96\pi t} \right)$$
