Physics Waves

The fastest any wave can travel is the speed of light in a vacuum. It is known that this speed is approximately 3 * 10 (to the power 8) metres per second (m/s). In physics, this speed is considered the upper limit for the speed of all electromagnetic waves. Common examples of it include light, radio waves, X-rays, and gamma rays. You should know that no wave in classical physics exceeds this speed. As Class 11 students, you know that the medium is always important to consider when calculating wave speed, it is not surprising to know that when electromagnetic waves pass through a medium, such as glass or water, their speed decreases due to interactions with the medium's particles.

To know how fast a wave travels, it is important to know through which medium it pass through. The wave velocity or the speed at which a wave travels shows how quickly the energy moves through a specific medium. It could be air, water, or steel. We calculate waves speed in physics using the formula:
Wave speed = Frequency * Wavelength,
That is,
$v=f \lambda$
This tells us that the speed depends on how often the wave oscillates. It also shows us the distance between its crests or troughs. The speed of a wave is medium-dependent because of particle density and elasticity of the medium. For instance, the speed of sound waves is faster than in solids like steel. 

The equation of a travelling wave propagating along the positive y- direction is given by the displacement equation:

y (x, t) = a sin ( $\omega t-kx)$ ……… (i)

Linear mass density $\mu =$ 8.0 $*{10}^{-3}$ kg/m and frequency of the tuning fork,  $\nu$ = 256 Hz

Amplitude of the wave, a= 5.0 cm = 0.05 m …. (ii)

Mass of the pan, m = 90 kg and tension of the string, T = mg = 90 $*9.8$ = 882 N

The velocity of the transverse wave, v is given by the relation:

v = $\sqrt{\frac{T}{\mu }}$ = $\sqrt{\frac{882}{8.0\mathrm{}*{10}^{-3}}}$ = 332 m/s

Angular frequency,  $\omega$ = 2 $\pi \nu$ = 2 $\pi *256$ = 1608.5 rad/s = 1.6 $*{10}^3$ rad/s

Wavelength,  $\lambda$ = $\frac{v}{\nu }$ = $\frac{332}{256}$ = 1.297 m

Propagation constant, k = $\frac{2\pi }{\lambda }$ = $\frac{2\pi }{1.297}$ = 4.844 /m

Substituting these values in the displacement equation, we get

y (x, t) = a sin ( $\omega t-kx)$

y (x, t) = 0.05 sin (1.6 $*{10}^3$ t – 4.844 x) m

(a) The given harmonic wave is

y(x, t) = 7.5 sin (0.0050x +12t + $\pi$ /4)

For x = 1 cm and t = 1 s

y(1, 1) = 7.5 sin (0.0050 +12+ $\mathrm{\pi }$ /4)

= 7.5 sin( 12.0050 + $\frac{\pi }{4}$ )

= 7.5 sin (12.0050 + 0.7854)

= 7.5 sin (732.84 $°$ )

= 7.5 sin (90 $*8+$ 12.81) $°$

= 7.5 sin 12.81 $°$

=1.6629 cm

The velocity of the oscillation at a given point and the time is given as:

v = $\frac{d}{dt}$ y(x,t) = $\frac{d}{dt}\left[7.5\mathrm{}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{}(0.0050x+12t+\pi /4)\right]$

= 7.5 $*12\mathrm{c}\mathrm{o}\mathrm{s}?(0.0050x+12t+\pi /4$ )

At x= 1 cm and t= 1 s

v= y (1,1) = 90 cos (12.005 + $\frac{\pi }{4}$ ) = 90 cos(12.81 $°$ ) = 87.75 cm/s

Now the equation of a propagating wave is given by

Y(x, t) = a sin (kx + $\omega$ t + $\phi )$ , where

k = $\frac{2\pi }{\lambda }$ or $\lambda$ = $\frac{2\pi }{k}$ and $\omega =2\pi v$

Speed, v = $\frac{\omega }{k}$ , where $\omega$ = 12 rad/s and k = 0.0050 /m

v = $\frac{12}{0.0050}$ = 2400 cm/s

Hence, the velocity of the wave oscillation at x=1 cm and t=1s is not equal to the velocity of the wave propagation.

(b) Propagation constant is related to wavelength as:

$\lambda$ = $\frac{2\pi }{k}$ = $\frac{2\pi }{0.0050}$ = 1256 cm = 12.56 m

Therefore, all the points at distances n $\lambda$ (n = $\pm 1,\pm 2\dots \dots .\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{}\mathrm{o}\mathrm{n})$ i.e. $\pm 12.56m,\pm 25.12m\dots \dots .$ and so on for x=1 cm, will have the same displacement as the x=1 cm points at t= 2s, 5s and 11s.

(a) For the stationary observer:

Frequency of the sound produced by the whistle,  = 400 Hz

Speed of sound = 340 m/s

Velocity of wind, v = 10 m/s

As there is no relative motion between the source and the observer, the frequency of the sound heard by the observer will be the same as that produced by the source, i.e. 400 Hz. The wind is blowing towards the observer. Hence, the effective speed of the sound increases by 10 units, i.e.

Effective speed of the sound, $v_e$ = 340 + 10 = 350 m/s

The wavelength ( $\lambda )\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}$ heard by the observer is given by the relation:

$\lambda =\frac{v_e}{\nu }$ = $\frac{350}{400}$ = 0.875 m

(b) For the running Observer:

$\mathrm{V}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{b}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{e}\mathrm{r}$ , $v_o$ = 10 m/s

$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{b}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{e}\mathrm{r}$ is moving towards the source. As a result of the relative motions of the source and the observer, there is a change in frequency (f''')

$\mathrm{T}\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{b}\mathrm{y}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}:$ f''' = ( $\frac{v+v_o}{v})\nu$ = ( $\frac{340+10}{340})*400=$ 411.76 Hz

Since the air is still, the effective speed of sound = 340 + 0 = 340 m/s

The source is at rest. Therefore, the wavelength of the sound will not change, i.e. $\lambda$ remains 0.875 m.

Hence, the given two situations are not exactly identical.