Wave Motion and String Waves Concept Page - 3

Example

Phase difference in a travelling sound wave

Example: A wave of frequency 500 Hz$500Hz$ has a phase velocity of 360 m/s$360m/s$. What will be the phase difference between the two displacements at a certain point in a time interval of 10−3${10}^{-3}$ seconds?

Solution:
According to the concept of simple harmonic wave,
y=asin2πλ(vt−ϕ)$y=a\sin {\displaystyle \frac{2\pi }{\lambda }}(vt-\varphi )$
Now, phase angle at point x$x$ for timet1$t_1$ is given by:
ϕ1=2πλ(vt1−x)${\varphi }_1={\displaystyle \frac{2\pi }{\lambda }}(v{t}_1-x)$
Now for time t2$t_2$, we may write:
ϕ2=2πλ(vt2−x)${\varphi }_2={\displaystyle \frac{2\pi }{\lambda }}(v{t}_2-x)$
So, phase difference ,(ϕ2−ϕ1)=2πλ(vt2−x)−2πλ(vt1−x)$({\varphi }_2-{\varphi }_1)={\displaystyle \frac{2\pi }{\lambda }}(v{t}_2-x)-{\displaystyle \frac{2\pi }{\lambda }}(v{t}_1-x)$=2πλv(t2−t1)$={\displaystyle \frac{2\pi }{\lambda }}v(t_2-t_1)$=2πnλλ(t2−t1)$={\displaystyle \frac{2\pi n\lambda }{\lambda }}(t_2-t_1)$=2πn(t2−t1)$=2\pi n(t_2-t_1)$=2π×500×10−3$=2\pi \times 500\times {10}^{-3}$=π radian$=\pi radian$

Example

Number of waves between two given points

Wave number is number of waves per unit distance.

Example: A sound wave of frequency 500 Hz$500\text{Hz}$ covers a distance of 1000 m$1000\text{m}$ in 5 s$5\text{s}$ between points x${\displaystyle x}$ and y${\displaystyle y}$. Then find the number of waves between x${\displaystyle x}$ and y${\displaystyle y}$ .

Solution:
ν=500 Hz$\nu =500\text{Hz}$, x=1000 m$x=1000m$ and t=5 s$t=5\text{s}$
Velocity of sound  v=xt=200 m/s$v={\displaystyle \frac{x}{t}}=200\text{m/s}$
But v=νλ$v=\nu \lambda$
Hence, λ=25 m/s$\lambda ={\displaystyle \frac{2}{5}}\text{m/s}$
Now nλ=x or n=2500$n\lambda =x\text{or}n=2500$

Example

Angular Frequency

Some important terms for a wave are wave frequency, wave number, wave length, velocity etc. We are discussing about wave frequency and its units,  its mathematical formula and its properties.
Frequency is explained by taking an example of coil of a slinky which is moved with completing two cycles in one second. The rate of its motion is 2 cycles/second. This rate is referred to as the wave frequency. Thus the wave frequency shows that how much the medium particles undergo in vibration when a wave is passed through that medium. It's cycles per second or waves per second or vibrations per second. It is not similar to period.

Example: A whistle of frequency 540 Hz rotates in a  horizontal circle of radius 2m at an angular speed of 15 rad/s.  What is the highest frequency heard by a listener at rest with respect to the centre of circle?
(velocity of sound in air =330 ms−1$=330m{s}^{-1}$)

Solution:
Highest frequency at rest  f1=(vv−vs)f$f_1=({\displaystyle \frac{v}{v-v_s}})f$
Now, vs=rω=30 ms−1$v_s=r\omega =30m{s}^{-1}$
So, f1=(330300)×540=594 Hz$f_1=({\displaystyle \frac{330}{300}})\times 540=594Hz$

Definition

For a given wave function infer the different parameters of a wave

Example: One end of a long string of linear mass density 8.0×10−3 kgm−1$8.0\times {10}^{-3}kg{m}^{-1}$ is connected to an electrically driven tuning fork of frequency 256 Hz$256Hz$. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg$90kg$. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t=0$t=0$, the left end (fork end) of the string x=0$x=0$ has zero transverse displacement (y=0$y=0$) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm$5.0cm$. Write down the transverse displacement y$y$ as function of x$x$ and t$t$ that describes the wave on the string.

Solution:

The equation of a travelling wave propagating along the positive y-direction is given by the displacement equation: y(x,t)=Asin(wt−kx)$y(x,t)=A\sin (wt-kx)$  ...(i)
Linear mass density, μ=8.0×10−3 kgm−1$\mu =8.0\times {10}^{-3}kg{m}^{-1}$Frequency of the tuning fork, ν=256 Hz$\nu =256Hz$Amplitude of the wave, A=5.0 cm=0.05 m$A=5.0cm=0.05m$  ...(ii)
Mass of the pan, m=90 kg$m=90kg$ 
Tension in the string, T=mg=90×9.8=882N$T=mg=90\times 9.8=882N$The velocity of the transverse wave, v$v$, is given by the relation:
v=tμ−−√$v=\sqrt{{\displaystyle \frac{t}{\mu }}}$=8828.0×10−3=332 m/s$={\displaystyle \frac{882}{8.0\times {10}^{-3}}}=332m/s$
Angular Frequency, ω=2πν$\omega =2\pi \nu$=2×3.14×256$=2\times 3.14\times 256$=1608.5≈1.6×103rad/s$=1608.5\approx 1.6\times {10}^{3}rad/s$  ...(iii)
Wavelength, λ=vν=332256 m$\lambda ={\displaystyle \frac{v}{\nu }}={\displaystyle \frac{332}{256}}m$
∴$\therefore$ propagation constant, k=2πλ$k={\displaystyle \frac{2\pi }{\lambda }}$=2×3.14332256=4.84 m−1$={\displaystyle \frac{2\times 3.14}{{\displaystyle \frac{332}{256}}}}=4.84m^{-1}$  ...(iv)
Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get the displacement equation:
y(x,t)=0.05sin(1.6×103t−4.84x) m$y(x,t)=0.05\sin (1.6\times {10}^{3}t-4.84x)m$.

Example

Understand direction of movement of particles of a string in case of waves on string

Example: If phase of the particle A at time t is greater than phase of next particle B at that time. Find the direction of travel of wave.

Solution:
Wave travels towards higher phase. Or you can put in the equation Y=asin(−t+kx).$Y=asin(-t+kx).$ Put t=0 as we can take any time. We can see that the forward going wave goes from lower phase lets say 0 to higher phase, lets say π$\pi$ which is at higher x$x$. Therefore particle will move from B$B$ to A$A$ .

Example

Use force analysis to find formula for velocity of transverse wave on a string

A string of length 100 cm$100cm$  and mass  0.5 gm$0.5gm$ is streched with a force of 20 N$20N$. It is pulled at a distance of  12.5 cm$12.5cm$  from one end. Find the frequency of the note emitted?

Solution:
Tension(T) = 20N$20N$
μ=0.5×10−3100×10−2=5×10−4 kg/m$\mu ={\displaystyle \frac{0.5\times {10}^{-3}}{100\times {10}^{-2}}}=5\times {10}^{-4}kg/m$
n=12lTμ−−√$n={\displaystyle \frac{1}{2l}}\sqrt{{\displaystyle \frac{T}{\mu }}}$
l=12.5×10−2 m$l=12.5\times {10}^{-2}m$
n=12×12.5×10−2205×10−4−−−−−−−−√=4×200=800 Hz$n={\displaystyle \frac{1}{2\times 12.5\times {10}^{-2}}}\sqrt{{\displaystyle \frac{20}{5\times {10}^{-4}}}}=4\times 200=800Hz$

Example

Write equation of transverse sinusoidal wave on a string when initial conditions are given

Example: A 100 Hz$100Hz$ sinusoidal wave is travelling in the positive x-direction along a string with a linear mass density of 3.5×10−3kg/m$3.5\times {10}^{-3}kg/m$ and a tension of 35 N$35N$. At time t=0$t=0$, the point x=0$x=0$, has maximum displacement in the positive y direction. Next when this point has zero displacement the slope of the string is π/20$\pi /20$. Find the expression that represent (s) the displacement of string as a function of x$x$ (in metre) and t$t$ (in second).

Solution:
Let the wave have the form y=Asin(ωt−kx+ϕ)$y=Asin(\omega t-kx+\varphi )$
Since the frequency is 100Hz$100Hz$, ω=2πν=200π$\omega =2\pi \nu =200\pi$,
speed of the wave=Tμ−−√=ωk$\sqrt{{\displaystyle \frac{T}{\mu }}}={\displaystyle \frac{\omega }{k}}$⟹k=2π$⟹k=2\pi$
Since displacement is maximum at (x,t)=(0,0)$(x,t)=(0,0)$, sin(0+0+ϕ)=1$sin(0+0+\varphi )=1$⟹ϕ=π2$⟹\varphi ={\displaystyle \frac{\pi }{2}}$
Thus the wave is y=Acos(ωt−kx)$y=Acos(\omega t-kx)$
Slope=∣∣∣dydx∣∣∣=Aksin(ωt−kx)=π20$Slope=\left|{\displaystyle \frac{dy}{dx}}\right|=Aksin(\omega t-kx)={\displaystyle \frac{\pi }{20}}$ at (x,t)=(0,0)$(x,t)=(0,0)$
Thus Ak=π20$Ak={\displaystyle \frac{\pi }{20}}$⟹A=0.025 m$⟹A=0.025m$Final equation is:
y=0.025cos(200πt−2πx)$y=0.025cos(200\pi t-2\pi x)$

Definition

Particle Velocity in waves

Particle velocity is the velocity of a particle (real or imagined) in a medium as it transmits a wave. The SI unit of particle velocity is the metre per second (m/s). In many cases this is a longitudinal wave of pressure as with sound, but it can also be a transverse wave as with the vibration of a taut string.

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