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Relation Between Particle Velocity And Wave Velocity

Relation Between Particle Velocity And Wave Velocity :- In wave motion, it is important to distinguish between the motion of individual particles of the medium and the propagation of the wave itself. The velocity with which the wave travels through the medium is called wave velocity, whereas the velocity with which the particles of the medium vibrate about their mean positions is called particle velocity. Though both occur simultaneously in wave motion, they are fundamentally different in nature.

Wave velocity is the speed at which the disturbance or wave energy propagates through the medium. It depends on the physical properties of the medium, such as elasticity and inertia, and remains constant for a given medium. Wave velocity is given by

and it is the same for all particles of the medium.

On the other hand, particle velocity is the velocity of oscillation of a particle about its equilibrium position. In a plane progressive harmonic wave, each particle executes simple harmonic motion. Therefore, particle velocity varies continuously with time and position. It is maximum at the mean position and zero at the extreme positions. The maximum particle velocity is given by

A key point of difference is that wave velocity is the same for all particles, while particle velocity is different for different particles at the same instant. Also, wave velocity represents the transport of energy, whereas particle velocity represents only the local oscillatory motion of the medium’s particles. Importantly, particles do not travel along with the wave; they merely oscillate about their mean positions while the wave advances forward.

Thus, although particle velocity and wave velocity are related through wave motion, they are entirely different physical quantities and should not be confused with each other.

Consider a plane progressive harmonic wave travelling along the positive -direction given by

$\displaystyle y=Asin(\omega t-kx)$ …..(1)

Particle velocity is given by the time derivative of displacement :

$\displaystyle {{\upsilon }_{p}}=\frac{\partial y}{\partial t}=\frac{\partial }{\partial t}\left[ Asin(\omega t-kx) \right]=A\omega cos(\omega t-kx)$ …..(2)

Wave velocity is given by

$\displaystyle v=\frac{\lambda }{T}=\frac{\lambda }{{}^{2\pi }\!\!\diagup\!\!{}_{\omega }\;}=\frac{\omega }{{}^{2\pi }\!\!\diagup\!\!{}_{\lambda }\;}=\frac{\omega }{k}$ …..(3)

The Slope of displacement curve or strain (defined as the rate of change of displacement with position) is given by

$\displaystyle \frac{\partial y}{\partial x}=\frac{\partial }{\partial x}\left[ Asin(\omega t-kx) \right]=-Akcos(\omega t-kx)$ …..(4)

Dividing particle velocity by slope, i.e., dividing equation (2) by (4) :

$\displaystyle \frac{\left( \frac{\partial y}{\partial t} \right)}{\left( \frac{\partial y}{\partial x} \right)}=\frac{A\omega cos(\omega t-kx)}{-Akcos(\omega t-kx)}=-\frac{\omega }{k}=wave\text{ }velocity=v$

$\displaystyle \therefore {{v}_{p}}=-v\left( \frac{\partial y}{\partial x} \right)$ …..(5)

⇒ Particle velocity = – (Wave velocity) × slope of the wave

From equation (5) we note that :

The slope tells how rapidly displacement changes with position.
The particle velocity is proportional to the slope of the wave at that point.
Where slope is maximum → particle speed is maximum.
Where slope is zero → particle velocity is zero.
The negative sign shows that particle motion is opposite to the direction of increasing at that instant.

Particle acceleration

The particle acceleration is defined as the rate of change of particle velocity with time or the second partial derivative of displacement y (x,t) with respect to time t.

Differentiating equation (2) with respect to time,

$\displaystyle {{a}_{p}}=\frac{\partial }{\partial t}\left( {{\upsilon }_{p}} \right)=\frac{\partial }{\partial t}\left[ A\omega cos(\omega t-kx) \right]$

$\displaystyle \therefore {{a}_{p}}=-{{\omega }^{2}}Asin(\omega t-kx)=-{{\omega }^{2}}y(x,t)$ …..(6)

Thus, the particle’s acceleration is proportional to its displacement and is equal to times the displacement, as in simple harmonic motion.

The figure below illustrates the particle velocity (v_p) and particle acceleration (a_p), as given by equations (5) and (6), for two points A and B on a string when a sinusoidal wave propagates along the positive x-direction.

At point A, the slope of the curve is positive; therefore, according to equation (5), the particle velocity (v_p) is negative and downward. As the displacement is positive, equation (6) implies that the particle acceleration (a_p) is also negative and downward.

For point B, the slope of the curve is negative. Hence, from equation (5), the particle velocity (v_p) is positive and directed upward. Since the displacement of the particle is positive, equation (6) shows that the particle acceleration (a_p) is negative and directed downward.

Example 1.

A transverse wave is travelling along a string from left to right. The adjoining figure represents the shape of the string at a given instant.

At this instant, among the following, choose the wrong statement :-

(A) Points D, E and F have upwards positive velocity.

(B) Points A, B, H and I have downwards negative velocity.

(D) Points A, E and I have minimum velocity.

Solution :

We know that

Particle velocity = – (Wave velocity) × slope of the wave

(A) For upward positive velocity, $v_{p} = + v e$ , the so slope must negative, which is at the points $D, E$ and $F.$

(B) For downwards negative velocity, $v_{p} = - v e$ , the so slope must positive, which is at the points A, B, H and I $.$

(D) For maximum magnitude of velocity, $∣$ slope $∣=$ maximum, which is at $A, E$ and $I$ .
Hence, option (D) is wrong which is the required option.

Note :

The shape of the string after a small time interval $Δt\Delta t$ is shown by the dotted curve. The distances moved by different points of the string during this time are indicated by arrows. Hence, the length of each arrow represents the magnitude of the velocity of the corresponding point. This diagram once again confirms the discussion given above.

Example 2.

The equation of a wave is $\displaystyle y(x,t)=0.05\sin \left[ \frac{\pi }{2}(10x-40t)-\frac{\pi }{4} \right]m$ .

Find :

(a) The wavelength, the frequency and the wave velocity.

(b) The particle velocity and acceleration at x = 0.5 m and t = 0.05 s.

Solution :

The equation may be rewritten as,

$\displaystyle y(x,t)=0.05\sin \left( 5\pi x-20\pi t-\frac{\pi }{4} \right)m$

Comparing this with Equation of a Plane Progressive Wave, $\displaystyle y=Asin\left( \omega t-kx\pm \phi \right)$ , we have

$\displaystyle k=\frac{2\pi }{\lambda }=5\pi \,rad/m$

$\displaystyle \Rightarrow \lambda =0.4m$

The angular frequency is,

$\displaystyle \omega =2\pi f=20\pi \,rad/s$

$\displaystyle \Rightarrow \nu =10Hz$

The wave velocity is,

$\displaystyle \upsilon =\frac{\omega }{k}=\nu \lambda =10\times 0.4=4m{{s}^{-1}}$ in positive x-direction.

(b) The particle velocity and acceleration are,

$\displaystyle {{\upsilon }_{p}}=\frac{\partial y}{\partial t}=-(20\pi )(0.05)\cos \left( 5\pi \times 0.5-20\pi \times 0.05-\frac{\pi }{4} \right)=2.22m/s$

$\displaystyle {{a}_{p}}=\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}=-{{(20\pi )}^{2}}(0.05)\sin \left( 5\pi \times 0.5-20\pi \times 0.05-\frac{\pi }{4} \right)\approx 140m/{{s}^{2}}$

Relation Between Particle Velocity And Wave Velocity

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