WAVES
Essay Question Answers
Q1. Explain theformation of stationary waves in stretched strings and hence deduce thelaws of transverse waves in stretched strings.
Ans. When the stretched string is plucked at the middle transverse waves are generated these transverse waves get reflected at its ends.These reflected waves travel in opposite directions along the length of the string they combine. together to form stationary waves with nodes at ends.
Explanation: Two transverse progressive waves with displacements $Y_{1},\:Y_{2}$ are
$$Y_{_1}=\:a\:sin\:(kx\:-\:\omega t)$$ $$ Y_{_2}=\:a\:sin\:(kx \:+\:\omega t)$$
According to principle of superposition,.
$$Y=Y_{1}+Y_{2} $$ $$\begin{aligned}&Y=a\sin\left(kx\:-\:\omega t\right)+a\sin\left(kx \:+\:\omega t \right)\\&Y=a\left(\sin\left(kx\:-\:\omega t\right)+\sin\left(kx \:+\:\omega t\right)\right)\\&[\because sin(A+B)\:+\:sin(A-B)\:=\:2\:sin(A)\:cos(B)]\end{aligned}$$ $$\therefore \:\:Y=2a \sin(kx) \cos(\omega t) $$
In the above equation $2a \sin( kx)$ is shows the amplitude of a stationary wave..
- When $x=\frac{n\lambda}{2}$ $(\mathbf{n}=0,1,2,3,4,5,.......)$ the amplitude becomes zero. These positions of zero amplitude are known as Nodes.( $Y=0$ )
- When $x=(2n+1)\frac{\lambda}{4}$ ( $\mathbf{n}=0,1,2,3,4,5,.......)$ the amplitude becomes maximum (2a). These positions of 2a amplitude are known as Antinodes. $ Y= 2a$
Expression to the frequency of transverse wave
1) If $L$ is the length of the string, then the velocity of transverse wave $(V)$ travelling along engthof the stretched string is given by $V=\sqrt{\frac{T}{\mu}}$ . $T$ is the tension applied to the string, $\mu$ is the mass per unit length of the string or linear density. The distance between adjacent nodes is so that in string fixed at both ends there must be an integrated number. $\mathbf{n}$ of half wave lengths.
$$L=\frac{n\lambda}{2}$$
Wave length $\lambda=\frac{2L}{n}$
Frequency $\nu_{_n}=\frac{V}{\lambda}$
$$\nu_{_n}=\frac{n}{2L}\sqrt{\frac{T}{\mu}}$$
$n=1$ then frst harmonic $\nu_{_1}=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$
$n=2$ then second harmonic $\nu_{_2}=\frac{2}{2L}\sqrt{\frac{T}{\mu}}=2\mathbf{\nu}_{_1}$
$n=3$ then third harmonic $\nu_{_3}=\frac{3}{2L}\sqrt{\frac{T}{\mu}}=3\mathbf{\nu}_{_1}$
$$\nu_{_1}:\mathrm{\nu}_{_2}:\nu_{_3}=1:2:3$$
Transverse Laws :
$1^{st}$ Law : The fundamental frequency of vibrating string is inversely proportional to thelength of the stringwhen tension andlineardensity are constant.
$$\nu\alpha\frac{1}{L} \:\:( \mathbf{T},\mathbf{\mu} \:\:are \:\:constant)$$
$2^{nd}$ Law: The fundamental frequency of vibrating string is directly proportional to the square root of the tension when length of the string and linear density are constant.
$$\nu\alpha\sqrt{T}\:\:(L,\mu\:\: are \:\:constant)$$
$3^{rd}$ Law: $\vdots$ The fundamental frequency of vibrating string is inversely proportional to the square root of linear density of the string and tension are constant.
$$\nu\alpha\sqrt{\frac{1}{\mu}}\:\:(L,T\:\:are\:\:constant)$$
Q2. Explain the formation of stationary waves in an air column enclosed in an open pipe. Derive the equations for the frequencies of the harmonics produced
Ans. Formation of stationary waves in open pipe:- If both ends of the tube are opened thenit is known as an open organpipe.When a sound waveis sent through a pipe from one open end it gets reflected from the other open end due topressure difference.Anti nodes are formed at the two openends and a node is between them. Frequencies of Harmonicsin open pipes:
In the open pipe,while vibrating the length of the air column $L$ is equal to the multiple of integer number $n$ of ahalf wavelength
$$L=\frac{n\lambda}{2}$$
Wave length $\lambda=\frac{2L}{n}$
Frequency $\nu_{_n}=\frac{V}{\lambda}$
$$\nu_{_n}=\frac{nV}{2L} $$
$n=1$ then frst harmonic $\nu_{_1}=\frac {V}{2L} $
$n=2$ then second harmonic $\nu_{_2}=\frac{2V}{2L} =2\mathbf{\nu}_{_1}$
$n=3$ then third harmonic $\nu_{_3}=\frac{3V}{2L} =3\mathbf{\nu}_{_1}$
$$\nu_{_1}:\mathrm{\nu}_{_2}:\nu_{_3}=1:2:3$$
Q3. How are stationary waves formed in closed pipes? Explain the various modes of vibrations and obtain relations for their frequencies.
Ans. Formation of stationary waves in a closed pipe:-. If one end of a pipe is closed and another end is opened,thatpipe is called a closed pipe.
Explanation: When a sound wave is sent through an closed organ pipe the wave gets reflected at the end of the pipe. These incident and reflected waves having same frequency travelling in the opposite directions are superimposed along the length of thepipe andform alongitudinal stationary wave.At theopen end always an antinode and at the closed end always node is formed
Frequencies of Harmonics in closed pipes: In the closed pipe,while vibrating the length of the air column‘L'is equal to the
multiple of odd number $(2\mathbf{n}+1)\frac{\lambda}{4}$ of a quarter of wavelength
$$L=(2\mathbf{n}+1)\frac{\lambda}{4}$$
Wave length $\lambda=\frac{4L}{(2\mathbf{n}+1)}$
Frequency $\nu_{_n}=\frac{V}{\lambda}$
$$\nu_{_n}=(2\mathbf{n}+1)\frac{V}{4L} $$
$n=0$ then fundamental frequency $\nu_{_{\circ}}=\frac {V}{4L} $
$n=1$ then first Overtone $\nu_{_1}=\frac{3V}{4L} =2\mathbf{\nu}_{_{\circ}}$
$n=2$ thenSecond Overtone $\nu_{_2}=\frac{5V}{4L} =3\mathbf{\nu}_{_{\circ}}$
$$\nu_{_{\circ}}:\mathrm{\nu}_{_1}:\nu_{_2}=1:2:3$$
Q4. What are beats? Obtain an expression for the beat frequency. Where and how are beats made use of ?
Ans. Beats areproduced due tointerference of sound waves.When two sounds of nearly equal frequency are superposed they will create a waxing and waning intensity of sound. This effect is called beats.
$$\text{Beat frequency} \:\: \nu=\nu_{_1} - \nu_{_2} $$
Let us consider two sound wave $Y_{1}$ and $Y_{2}$ of nearly equal frequency $\omega_{1}$ and $\omega_{2}$ each of amplitude a superpose each other then the resultant wave is given by
$$\begin{array}{rcl}Y_1=&a\cos{(\omega_1t)}\\Y_2=&a\cos{(\omega_2t)}\end{array}$$
According to principle of superposition, $Y=Y_{1}+Y_{2}$
$$\begin{gathered}
Y\:=\:a\:cos(\omega_{1}t)\:+\:a\:cos(\omega_{2}t) \\
Y\:=\:2a\:cos(\frac{\omega_{1}-\omega_{2}}{2})t\:Cos(\frac{\omega_{1}+\omega_{2}}{2})t \\
Y\:=\:2a\:cos\:2\pi(\frac{\nu_{_1}-\nu_{_2}}{2})t\:Cos\:2\pi(\frac{\nu_{_1}+\nu_{_2}}{2})t \\
\end{gathered}$$
In the above equation $2a \:\cos2\pi(\frac{\nu_{_1}-\nu_{_2}}{2})t$ it shows the amplitude of beats.
The intensity of the sound will be maximum when $2a \:\cos2\pi(\frac{\nu_{_1}-\nu_{_2}}{2})t$ is maximum
$$\therefore \:\: \nu=\nu_{_1}\:-\:\nu_{_2}$$
Importance of Beats:
- Beats can be used in tuning musical instruments.
- Beats are used in detecting dangerous gases in mines.
- Beats are used to produce special effects in cinematography
- Beats can be used to determine the unknown frequency of a tuning fork.
- Beats are used in heterodyne receivers range.
Q5. What is Doppler effect ? Obtain an expression for the apparent frequency of sound heard when the source is in motion with respect to an observer at rest.
Ans. Doppler effect: The apparent change in the frequency heard by an observer due to relative motion of source of sound and observer is known as Doppler effect.
Case (1) : Consider observer at rest and a source of sound $S$ moving with velocity ' $V_{s}$ away from an observer. Let the speed of sound in air $V$ , frequency $\nu_{_{\circ}}$ and time period $T_{_{\circ}}$ .As shown in the figure at the time $t=0$ source is at distance $L $ from the observer .
Therefore. At $t=0$ first crest reaches the observer at time $$t_{1}=\frac{L}{V}$$ $t=T_{o}$ second crest reaches the observer at time $$t_{2}=T_{ {\circ}}+\frac{L+V_{s}T_{o}}{V}$$ $t= nT_{o}$, $( \mathbf{n} + 1)$ crest reaches the observer at time $$t_{n+1}=nT_{\:o}+\frac{L+nV_{\:s}T_{\:o}}{\:V} $$
Time interval $$\begin{gathered}
\:t_{_{n+1}}-t_{_1}=\:nT_{_o}+\biggl(\frac{L+\:nV_{_s}T_{_o}}{V}\biggr )-\biggl(\frac{L}{V}\biggr ) \\
nT\:=\:nT_{_o}\left(1\:+\:\frac{V_{_s}}{V}\right) \\
T=T_{o}\left({1+\frac{V_{_s}}{V}}\right) \\
\mathrm{apparent\:\: frequency}\: \mathbf{\nu}={\frac{1}{T}}={\frac{1}{T_{o}\left({1+\frac{V_{_s}}{V}}\right)}} \\
\nu\:=\:\nu_{o}\biggl(1\:-\:\frac{V_{o}}{V}\biggr) \\
\end{gathered}$$
The above equation represents the apparent frequency while the source of sound is moving away from an observer.
Case (2): when source of sound is approaching the observer we replaced $V_{s}$ by $-\:V_{s}$ in the above equation
$\therefore$ apparent frequeny$$\nu^{\prime}\:=\:\nu_{o}\biggl(1\:+\:\frac{V_{s}}{V}\biggr)$$
Q6. What is Doppler shift ? Obtain an expression for the apparent frequency of sound heard when the observer is in motion with respect to a source at rest.
Ans. Doppler shift : The difference between actual frequency and apparent frequency is known as Doppler shift .
Case(1):Consider Source of sound at rest and Observer $O$ approaching source of sound $S$ with avelocity' $V_{o}$ .Let the speed of sound in air. $v$ frequency $\nu_{o}$ and time period $T_{o}$ .As shown in the figure at the time $t=0$ source is at distance $L $ from the observer .
Therefore. At $t=0$ first crest reaches the observer at time $$t_{1}=\frac{L}{V+V_{o}}$$ $t=T_{o}$ second crest reaches the observer at time $$t_{2}=T_{o}+\biggl(\frac{L-V_{o}T_{o}}{V+V_{o}}\biggr )$$ $t=nT_{o}, \text{then}\left(\mathbf{n}+1\right)$ crest reachs the observer at tiime $$t_{n+ 1}=nT_{o}+\biggl(\frac {L- nV_{s}T_{o}}{V+ V_{o}}\biggr )$$ Time interval
$$\begin{gathered}
\:t_{_{n+1}}-t_{_1}=\:nT_{_o}+\biggl(\frac{L-\:nV_{_o}T_{_o}}{V+V_{_o}}\biggr )-\biggl(\frac{L}{V+V_{_o}}\biggr ) \\
nT\:=\:nT_{_o}\left(1\:-\:\frac{V_{_o}}{V+V_{_o}}\right) \\
T=T_{o}\left({\frac{V}{V+V_{_o}}}\right) \\
\mathrm{apparent\:\: frequency}\: \mathbf{\nu}={\frac{1}{T}}={\frac{1}{T_{o}\left({\frac{V}{V+V_{o}}}\right)}} \\
\nu\:=\:\nu_{o}\biggl(1\:+\:\frac{V_{o}}{V}\biggr) \\
\end{gathered}$$
The above equation represents the apparent frequency while an observer approaching the source of sound.
Case (2): when Observer moving away from the source of sound we replaced $V_{\circ}$ by $-\:V_{\circ}$in the above equation$$\nu^{\prime}\:=\:\nu_{o}\biggl(1\:-\:\frac{V_{o}}{V}\biggr)$$