Wave Motion and String Waves Concept Page - 14

Definition

Describe resonance in vibrations of a sonometer wire with given external frequency

When the frequency of the applied force is equal to the natural frequency of the sonometer, the body vibrates with very large amplitude. Corresponding intensity of sound will be maximum. This phenomenon is known as resonance.

Example

Variation in velocity of a wave on a string on tension variation

A wire of uniform cross section is stretched between two points, 1 m apart. The wire is fixed at one end and a weight of 9 kg is hung over a pulley at the other end. It produces a fundamental frequency of 750 Hz. If the suspended weight is submerged in a liquid of density 5/9 that of the weight then find the velocity of the wave propagated along the wire.In case of fundamental vibration of the string
λ/2=L$\lambda /2=L$
or
λ=2L=2×1=2 m$\lambda =2L=2\times 1=2m$
As v=fλ$v=f\lambda$ and f=750 Hz$f=750Hz$,
we get vA=2×750=1500 m/s$v_A=2\times 750=1500m/s$
Now in case of wire under tension
vAvB=TATB−−−√${\displaystyle \frac{v_A}{v_B}}=\sqrt{{\displaystyle \frac{T_A}{T_B}}}$
or vB=1500TBTA−−−√$v_B=1500\sqrt{{\displaystyle \frac{T_B}{T_A}}}$
or vB=1500Mg′Mg−−−−√=1500g(1−σρ)g−−−−−−−−⎷=15001−59−−−−−√=1000 m/s$v_B=1500\sqrt{{\displaystyle \frac{M{g}^{\prime }}{Mg}}}=1500\sqrt{{\displaystyle \frac{g(1-{\displaystyle \frac{\sigma }{\rho }})}{g}}}=1500\sqrt{1-{\displaystyle \frac{5}{9}}}=1000m/s$

Example

Find time taken by a pulse formed in a string to travel a given length

Example: A string of length  l  hangs freely from a rigid support. Find the time required by a transverse pulse to travel from bottom to top of the string?

Solution:
f=v2l=T/m/l−−−−−√2l=12Tml−−−√=12mgml−−−√$f={\displaystyle \frac{v}{2l}}={\displaystyle \frac{\sqrt{{}^T{/}_{m/l}}}{2l}}={\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{T}{ml}}}={\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{mg}{ml}}}$
=12gl−−√$={\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{g}{l}}}$
Time taken =1f=2lg−−√.$={\displaystyle \frac{1}{f}}=2\sqrt{{\displaystyle \frac{l}{g}}}.$

Definition

Melde's Experiment

The experiment Melde by is a scientific experiment carried out in 1859 by the German physicist Franz Melde on the standing waves produced in a tense cable originally set oscillating by a tuning fork, later improved with connection to an electric vibrator. This experiment, "a lecture-room standby", attempted to demonstrate that mechanical waves undergo interference phenomena. Mechanical waves traveled in opposite directions form immobile points, called nodes. These waves were called standing waves by Melde since the position of the nodes and loops (points where the cord vibrated) stayed static.

A string undergoing transverse vibration illustrates many features common to all vibrating acoustic systems, whether these are the vibrations of a guitar string or the standing wave nodes in a studio monitoring room. In this experiment the change in frequency produced when the tension is increased in the string similar to the change in pitch when a guitar string is tuned will be measured. From this the mass per unit length of the string / wire can be derived.

Finding the mass per unit length of a piece of string is also possible by using a simpler method a ruler and some scales and this will be used to check the results and offer a comparison.

Example

Melde's Experiment

Principle of Melde's Experiment: A string undergoing transverse vibration illustrates many features common to all vibrating acoustic systems, whether these are the vibrations of a guitar string or the standing wave nodes in a studio monitoring room. In this experiment the change in frequency produced when the tension is increased in the string similar to the change in pitch when a guitar string is tuned will be measured. From this the mass per unit length of the string / wire can be derived. Finding the mass per unit length of a piece of string is also possible by using a simpler method a ruler and some scales and this will be used to check the results and offer a comparison.

Example: In Melde's experiment, it was found that the string vibrates in 5$5$ loops. When 10 gm.wt$10gm.wt$ is placed in a light pan. What weight must be placed to the pan to make it vibrates in 10$10$ loops? Neglect the weight of the pan.

Solution:
When a string of linear density m$m$, under tension T$T$, vibrates in n$n$ loops, the frequency, n2lTm−−−√=N${\displaystyle \frac{n}{2l}}\sqrt{{\displaystyle \frac{T}{m}}}=N$
or, Tn2=4l2N2m=constant$T{n}^2=4l^2{N}^{2}m=constant$T1n21=T2n22⇒T2=n21n12×T1=52102×10 gm.wt.=2.5 gm.wt.$T_1{n}_1^2=T_2{n}_2^2\Rightarrow T_2={\displaystyle \frac{n_1^2}{n_2^1}}\times T_1={\displaystyle \frac{5^2}{{10}^2}}\times 10gm.wt.=2.5gm.wt.$

Definition

State and use the Law of Length from the Laws of Transverse Vibrations of a String

Law of length: The fundamental frequency is inversely proportional to the resonating length (L) of the string. So,

f∝1L$f\propto {\displaystyle \frac{1}{L}}$

Definition

State and use the Law of Tension from the Laws of Transverse Vibrations of a String

Law of Tension: The fundamental frequency is directly proportional to the square root of the tension. So,

f∝T√$f\propto \sqrt{T}$

Definition

State and use the Law of Mass from the Laws of Transverse Vibrations of a String

Law of mass: The fundamental frequency is inversely proportional to the square root of the mass per unit length.

f∝1μ√$f\propto {\displaystyle \frac{1}{\sqrt{\mu }}}$

Definition

Amplitude of Standing Wave

Example: A string 120 cm in length sustains a standing wave, with the points of string at which the displacement amplitude is equal to  2√$\sqrt{2}$ mm being separated by 15.0 cm. Find the maximum displacement amplitude?

Solution:
From figure points A, B, C, D, E and F are having equal displacement amplitude.
Further, xE−xA−λ=4×15=60cm$x_E-x_A-\lambda =4\times 15=60cm$
As λ=2ln=2×120n=60;(n=0,1,2,3....)$\lambda ={\displaystyle \frac{2l}{n}}={\displaystyle \frac{2\times 120}{n}}=60;(n=0,1,2,\mathrm{3....})$
∴n=2×1206=4$\therefore n={\displaystyle \frac{2\times 120}{6}}=4$
So, it corresponds to 4th harmonic.
Also, distance of node from A=7.5cm$A=7.5cm$ as distance between B$B$ and C=15cm$C=15cm$ and node is between them. Taking node at origin, the amplitude of stationary wave can be written as
a=AsinKx$a=A\sin Kx$
Here, a=2√mm;K=2πλ=2π60$a=\sqrt{2}mm;K={\displaystyle \frac{2\pi }{\lambda }}={\displaystyle \frac{2\pi }{60}}$ and x=7.5cm$x=7.5cm$
∴2√=Asin(2π60×7.5)=Asinπ4=A×12√$\therefore \sqrt{2}=A\sin ({\displaystyle \frac{2\pi }{60}}\times 7.5)=A\sin {\displaystyle \frac{\pi }{4}}=A\times {\displaystyle \frac{1}{\sqrt{2}}}$
∴A=2mm$\therefore A=2mm$

Example

Understand and calculate velocity of a particle of standing wave at a fixed position

Example: A standing wave set up in a medium is y=4cos(πx3)sin40πt$y=4{\displaystyle \cos (\frac{\pi x}{3})\sin 40\pi t}$,  where x,y$x,y$ are in cm and t$t$ in sec. Find the velocity of medium particle at  x=3cm$x=3cm$ at t=18sec$t={\displaystyle \frac{1}{8}sec}$ .

Solution:
Velocity of medium particle v=dydt$v={\displaystyle \frac{dy}{dt}}$=4(40π) cos(πx3) cos(40πt)$=4(40\pi )cos\left({\displaystyle \frac{\pi x}{3}}\right)cos\left(40\pi t\right)$=160πcos(πx3) cos(40πt)$=160\pi cos\left({\displaystyle \frac{\pi x}{3}}\right)cos\left(40\pi t\right)$
At x=3cm, t=18s$x=3cm,t={\displaystyle \frac{1}{8}}s$
v=160π cos(π) cos(5π)$v=160\pi cos(\pi )cos(5\pi )$=160π (−1) (−1)$=160\pi (-1)(-1)$=160π cm/s

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