Wave Motion and String Waves Concept Page - 6

Example

Use power delivered at a point and total power delivered during one time period in a travelling sound wave

Example: A stretched rope having linear mass density 5×10−2kgm−1$5\times {10}^{-2}kg{m}^{-1}$ is under a tension of 80 N. Find the power that has to be supplied to the rope to generate harmonic waves at a frequency of 60Hz and an amplitude of 6 cm.

Solution:
Power P=2π2A2ν2Vμ$P=2{\pi }^2{A}^2{\nu }^{2}V\mu$
=2×(3.14)2×(6×10−2)2×(60)2×Tμ−−√×μ$=2\times (3.14{)}^2\times (6\times {10}^{-2}{)}^2\times (60{)}^2\times \sqrt{\frac{T}{\mu }}\times \mu$
=2×(3.14)2×(6×10−2)2×(60)2×5×10−2×80−−−−−−−−−−−√$=2\times (3.14{)}^2\times (6\times {10}^{-2}{)}^2\times (60{)}^2\times \sqrt{5\times {10}^{-2}\times 80}$
=10.368×103$=10.368\times {10}^3$
=10368 W$=10368W$

Example

State the principle of superposition of waves

The principle of superposition may be applied to waves whenever two (or more) waves traveling through the same medium at the same time. The waves pass through each other without being disturbed. According to this principle net displacement of the medium at any point in space or time, is simply the sum of the individual wave displacements.

Basic Assumption is:

The medium is non-dispersive (all frequencies travel at the same speed) since the Gaussian wave pulses do not change their shape as they propagate. If the medium was dispersive, then the waves would change their shape.

Example

Find the amplitude of resultant of two superimposing waves with same frequency and use in problems

Example: Two light waves are represented by y1=4sinωt$y_1=4sin\omega t$ and y2=3sin(ωt+π2)$y_2=3sin(\omega t+{\displaystyle \frac{\pi }{2}})$ . What will be the resultant amplitude of superposition?

Solution: The two waves are y1=4$y_1=4$ sinωt$sin\omega t$ and y2=3$y_2=3$ cosωt$cos\omega t$, the resultant wave y=y1+y2$y=y_1+y_2$
=4$=4$ sinωt$sin\omega t$ + 3 cosωt$cos\omega t$.
The resultant amplitude =42+32−−−−−−√$=\sqrt{4^2+3^2}$
=25−−√=5cm$=\sqrt{25}=5cm$

Example

Addition of two superimposing waves using vector method

Example: When two progressive waves y1=4sin(2x −6t)$y_1=4\sin (2x-6t)$ and y2=3sin(2x−6t−π2)$y_2=3\sin (2x-6t-\frac{\pi }{2})$ are superimposed, Find the amplitude of the resultant wave.

Solution:
Resultant amplitude =A21+A22+2A1A2cosθ−−−−−−−−−−−−−−−−−√$=\sqrt{A_1^2+A_2^2+2A_1{A}_{2}cos\theta }$
phase difference =θ=π2$=\theta =\frac{\pi }{2}$
=42+32−−−−−−√$=\sqrt{4^2+3^2}$
=5cm$=5cm$

Definition

Principle of Superposition

The superposition principle, also known as superposition property, states that, for all linear systems, the net response at a given place and time caused by two or more stimuli is the sum of the responses which would have been caused by each stimulus individually.

Example

Find Pressure Amplitude at a given point

Example: Due to a point isotropic sound source, the intensity at a point is observed as 40 dB$40dB$. The density of air is ρ=(15/11) kg/m$\rho =(15/11)kg/m$3${}^3$ and velocity of sound in air is 330 m/s$330m/s$. Based on this information answer the following questions. Find the pressure amplitude at the observation point.

Solution:
Given:    L=40dB$L=40dB$
Density of air    ρ=1511kg/m3$\rho ={\displaystyle \frac{15}{11}}kg/m^3$
Velocity of sound in air         v=330m/s$v=330m/s$
Loudness of the sound           L=10log10IIo$L=10lo{g}_{10}{\displaystyle \frac{I}{I_o}}$
40=10log10I10−12$40=10{\log }_{10}{\displaystyle \frac{I}{{10}^{-12}}}$
⟹I=10−8$⟹I={10}^{-8}$Let Po$P_o$ be the pressure amplitude.
Now   I=P2o2ρv⟹P2o=2ρvI$I={\displaystyle \frac{P_o^2}{2\rho v}}⟹P_o^2=2\rho vI$
Thus   P2o=2×1511×(330)×10−8$P_o^2=2\times {\displaystyle \frac{15}{11}}\times (330)\times {10}^{-8}$
⟹Po=3×10−3N/m2$⟹P_o=3\times {10}^{-3}N/m^2$

Example

Principle of superposition to write the equation of resultant sound wave

Superposition of waves results in maximum and minimum of intensities such as in case of standing waves. This phenomenon is called as interference. Another type of superposition results in interference in time which is called as beats. In this case waves are analyzed at a fixed point as a function of time. If two waves are of nearly same frequency are superimposed, at a particular point, intensity of combined waves gives a periodic peak and fall. This phenomenon is beats. If ω1${\omega }_1$ and ω2${\omega }_2$ are the frequencies of two waves then by superimposed y=y1+y2$y=y_1+y_2$, we get at x=0,
y=[2Acos(ω1−ω22)t]sin[ω1+ω22].t$y=[2A\cos \left({\displaystyle \frac{{\omega }_1-{\omega }_2}{2}}\right)t]\sin \left[{\displaystyle \frac{{\omega }_1+{\omega }_2}{2}}\right].t$
Thus amplitude frequency  is small and fluctuates slowly. A beat, i.e, a maximum of intensity occurs, also intensity depends on square of amplitude. The beat frequency is given by: ωbeat=|ω1−ω2|${\omega }_{beat}=|{\omega }_1-{\omega }_2|$
Numbers of beats per second is called as beat frequency. A normal ear can detect only upto 15 Hz of frequency because of persistence of ear.
A tuning fork of unknown frequency makes 3 beats per second with a standard fork of frequency 384 Hz. The beat frequency decreases when wax is put on prongs of first fork and the frequency of this fork would be determined by:
Standard frequency of the tuning fork is 384 Hz, and it produces 3 beats per second, it means either 387 Hz, or 381 Hz are the possible frequencies. Now if wax is added on the first fork. the frequency of that fork will decrease. Since the beat frequency also decreases it means that the frequency of the unknown fork is 387 Hz.

Example

Using relation between path difference and phase difference

Example: What will be the path difference for two sound waves having a phase difference of 600${}^0$.

Solution:
The relation between phase difference and path difference is given by, phase difference =2πλ×$={\displaystyle \frac{2\pi }{\lambda }}\times$ path  difference
60o=π3=2πλ×${60}^o={\displaystyle \frac{\pi }{3}}={\displaystyle \frac{2\pi }{\lambda }}\times$ path difference
Hence, path difference =λ6$={\displaystyle \frac{\lambda }{6}}$

Example

Use principle of superposition to find total displacement at a given point

Example: Equations of two progressive waves at a certain point in a medium are given by  :
 y1=a(sinωt+θ1)$y_1=a(\sin \omega t+{\theta }_1)$ and
 y2=a(sinωt+θ2)$y_2=a(\sin \omega t+{\theta }_2)$.
If amplitude and time period of resultant wave formed by the superposition of these two waves are same as those of either wave, then find (θ1−θ2)$({\theta }_1-{\theta }_2)$

Solution:
y1=a sin(ωt+θ1)$y_1=asin(\omega t+{\theta }_1)$
y2=a sin(ωt+θ2)$y_2=asin(\omega t+{\theta }_2)$
y=y1+y2$y=y_1+y_2$
=a sin ωt cos θ1+a cos ωt sin θ1+a sin ωt cos θ2+a cos ωt sin θ2$=asin\omega tcos{\theta }_1+acos\omega tsin{\theta }_1+asin\omega tcos{\theta }_2+acos\omega tsin{\theta }_2$=a sin ωt(cos θ1+cos θ2)+a cos ωt(sin θ1+sin θ2)$=asin\omega t(cos{\theta }_1+cos{\theta }_2)+acos\omega t(sin{\theta }_1+sin{\theta }_2)$
So, for same amplitude and time period.
cos θ1+cos θ2=0$cos{\theta }_1+cos{\theta }_2=0$
and (sin θ1+sin θ2)max=1$(sin{\theta }_1+sin{\theta }_2{)}_{max}=1$
or, sin θ1+sin θ2=0$sin{\theta }_1+sin{\theta }_2=0$
and (cos θ1+cos θ2)max=1$(cos{\theta }_1+cos{\theta }_2{)}_{max}=1$
So, θ1−θ2=2π3.

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