Wave Motion and String Waves Concept Page - 4

Example

Particle velocity of a given travelling longitudinal wave

Example: The velocity of sound is  vs$v_s$ in air. If density of air is increased twice then find the new velocity of sound.

Solution:
v=γpϱ−−−√$v=\sqrt{{\displaystyle \frac{\gamma p}{\varrho }}}$
So, v∝1ϱ√$v\propto {\displaystyle \frac{1}{\sqrt{\varrho }}}$
or, v1v2=1/ϱ1−−√1/ϱ2−−√${\displaystyle \frac{v_1}{v_2}}={\displaystyle \frac{1/\sqrt{{\varrho }_1}}{1/\sqrt{{\varrho }_2}}}$
⇒v1v2=ϱ2ϱ1−−−√$\Rightarrow {\displaystyle \frac{v_1}{v_2}}=\sqrt{{\displaystyle \frac{{\varrho }_2}{{\varrho }_1}}}$
⇒vsv2=2ϱϱ−−−√$\Rightarrow {\displaystyle \frac{v_s}{v_2}}=\sqrt{{\displaystyle \frac{2\varrho }{\varrho }}}$
⇒v2=vs2√$\Rightarrow v_2={\displaystyle \frac{v_s}{\sqrt{2}}}$

Example

Solve Problems involving particle acceleration of a given travelling longitudinal wave

Example: A source S of sonic oscillations of frequency 2200 Hz$2200Hz$ and a receiver R are located on the x-axis. At the moment t=0$t=0$ the source starts moving away from the receiver with an acceleration 1.7 m/s2$1.7m/s^2$. The velocity of sound in air is 340 m/s$340m/s$. Find the speed of the source if the sound is received by this receiver after 21 s$21s$.

Solution:
The source starts from R and reaches P after 21 s$21s$.
Let T$T$ be the total time taken by source to go to P(t1)$(t_1)$ from R and the sound to go from P to R(t2sec)$(t_{2}sec)$.
So, T=t1+t2=21$T=t_1+t_2=21$
Now, s=$s=$ distance between R and P=12at21,$={\displaystyle \frac{1}{2}}a{t}_1^2,$ where a$a$ is the acceleration of the source.
s$s$ is also equal to vt2$v{t}_2$, where v$v$ is the velocity of sound.
So, s=12at21=vt2$s=\frac{1}{2}a{t}_1^2=v{t}_2$=340t2=340(21−t1)$=340t_2=340(21-t_1)$
⇒$\Rightarrow$ 12×1.7×t21=340(21−t1)${\displaystyle \frac{1}{2}}\times 1.7\times t_1^2=340(21-t_1)$
or, t21+400t1−8400=0$t_1^2+400t_1-8400=0$
or, (t1+420)(t1−20)=0$(t_1+420)(t_1-20)=0$
∴t1=20 s$\therefore t_1=20s$
So the source travels for 20 s$20s$ from receiver and attained a velocity:
v=u+at=0+1.7×20$v=u+at=0+1.7\times 20$=34 m/s$=34m/s$

Definition

Sinusoidal form of a wave travelling with a constant speed

y(x,t)=asin(kx−ωt+ϕ)$y(x,t)=asin(kx-\omega t+\varphi )$
where a is the amplitude of the wave
y is the displacement from mean position
x is the position of an element along the wave
ϕ$\varphi$ is the initial phase angle
k is the angular wave number

Example

Speed of a sinusoidally travelling wave

The displacement y$y$ of a wave travelling in the x-direction is given by y=10−4sin(600t−2x+π2)$y={10}^{-4}\sin (600t-2x+{\displaystyle \frac{\pi }{2}})$ metre. Where x$x$ is expressed in metre and t$t$ in second. Calculate the speed of the wave motion.y=10−4sin(600t−2x+π2)$y={10}^{-4}\sin (600t-2x+{\displaystyle \frac{\pi }{2}})$
y=Asin(ωt−kx+ϕ)$y=Asin(\omega t-kx+\varphi )$
ω=600$\omega =600$
k=2$k=2$
V=ωk$V={\displaystyle \frac{\omega }{k}}$
=300 m/s$=300m/s$

Example

Find particle acceleration of a given travelling transverse wave

Example: A transverse wave along a string is given by  y=2sin(2π(3t−x)+π4)$y=2{\displaystyle \sin (2\pi (3t-x)+\frac{\pi }{4})}$, where  x  and  y  are in cm and t is in second. What is the acceleration of a particle located at  X=4cm   at t=1 sec ?

Solution:
y=2 sin(6πt−2πx+π4)$y=2sin(6\pi t-2\pi x+\frac{\pi }{4})$
vp=dydt=12π cos(6πt−2πx+π4)$v_p=\frac{dy}{dt}=12\pi cos(6\pi t-2\pi x+\frac{\pi }{4})$

ap=dvpdt$a_p=\frac{d{v}_p}{dt}$=−72π2 sin(6πt−2πx+π4)$=-72{\pi }^{2}sin(6\pi t-2\pi x+\frac{\pi }{4})$
At x=4, t=1$x=4,t=1$
ap=−72π2 sin(6π−8π+π4)$a_p=-72{\pi }^{2}sin(6\pi -8\pi +\frac{\pi }{4})$=−72π2×12√$=-72{\pi }^2\times \frac{1}{\sqrt{2}}$=−362√π2 cm/sec2$=-36\sqrt{2}{\pi }^{2}cm/se{c}^2$

Example

Potential energy per unit length at a given point in a travellling string wave

Example: A particle of mass is executing oscillations about. the origin on the x-axis. Its potential energy is V(x)=k|x|3$V(x)=k|x{|}^3$, where k$k$ is a positive constant. If the amplitude of oscillation is a$a$, then find its time period T$T$ ?

Solution:
V(x)=k|x|3$V(x)=k|x{|}^3$
Since, F=−dV(x)dx=−3k|x|2${\displaystyle F=-{\displaystyle \frac{dV(x)}{dx}}=-3k|x{|}^2}$            ......(1)$(1)$
x=asin(ωt)$x=a\sin (\omega t)$
This equation always fits to the differential equation:
d2xdt2=−ω2x${\displaystyle {\displaystyle \frac{d^{2}x}{d{t}^2}}=-{\omega }^{2}x}$
or md2xdt=−mω2x${\displaystyle m{\displaystyle \frac{d^{2}x}{dt}}=-m{\omega }^{2}x}$
⇒F=−mω2x$\Rightarrow F=-m{\omega }^{2}x$.........(2)$(2)$
Equation (1)$(1)$ and (2)$(2)$ give:
−3k|x|2=−mω2x$-3k|x{|}^2=-m{\omega }^{2}x$
⇒ω=3kxm−−−−√=3kam−−−−√[sin(ωt)]1/2${\displaystyle \Rightarrow \omega =\sqrt{{\displaystyle \frac{3kx}{m}}}=\sqrt{{\displaystyle \frac{3ka}{m}}}[\sin (\omega t){]}^{1/2}}$
⇒ω∝a√${\displaystyle \Rightarrow \omega \propto \sqrt{a}}$
⇒T∝1a√${\displaystyle \Rightarrow T\propto {\displaystyle \frac{1}{\sqrt{a}}}}$

Example

Derive and find kinetic energy per unit length at a given point in a travelling string wave

A uniform string of length / is fixed at both ends such that tension T is produced in it. The string is excited to vibrate with maximum displacement amplitude ao$a_o$. The maximum kinetic energy of the string for its fundamental tone is given as a20π2Txl${\displaystyle {\displaystyle \frac{a_0^2{\pi }^{2}T}{xl}}}$. Find x$x$.

Solution:
Since tension on the string is T$T$ so the velocity of the wave should be v=Tm−−−√$v=\sqrt{{\displaystyle \frac{T}{m}}}$, where m$m$ is mass per unit length.
Since frequency is n=vλ$n={\displaystyle \frac{v}{\lambda }}$  there for the frequency of the fundamental tone should be n1=12lTm−−−√$n_1={\displaystyle \frac{1}{2l}}\sqrt{{\displaystyle \frac{T}{m}}}$
Now consider an elemental length of a string at a distance x$x$ of size dx$dx$.
So its mass is: dm=mdx$dm=mdx$
Its oscillation energy is given by 12(mdx)a2(2πn1)2${\displaystyle \frac{1}{2}}(mdx)a^2({2\pi n_1)}^2$ =a20π2T2l2sin2(2πx2l)dx$={\displaystyle \frac{a_0^2{\pi }^{2}T}{2l^2}}{\sin }^2({\displaystyle \frac{2\pi x}{2l}})dx$      
[n1=12lTm−−−√$n_1={\displaystyle \frac{1}{2l}}\sqrt{{\displaystyle \frac{T}{m}}}$ ]
Integrating this we get, total energy as a20π2T4l${\displaystyle \frac{a_0^2{\pi }^{2}T}{4l}}$
This gives us x=4$x=4$

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