Physics Waves

The equation for a travelling harmonic wave is given by

y(x, t) = 2.0 cos 2 $\pi$ (10t – 0.0080 x + 0.35) or

= 2.0 cos (20 $\pi$ t – 0.016 $\pi$ x + 0.70 $\pi$ )

Comparing with classical equation y (x, t) = a $\sin ?(\omega t+kx+\phi )$ , we get

Amplitude a = 2.0 cm. Propagation constant, k = 0.016 $\pi$ , angular frequent, $\omega =20\pi rad/s$

From the relation $\phi =kx=\frac{2\pi }{\lambda }x$

(a) For x = 4 m = 400 cm, we have $\phi$ = 0.016 $\pi *400$ = 6.4 $\pi$ rad

(b) For x = 0.5 m = 50 cm, we have $\phi$ = 0.016 $\pi *50$ = 0.8 $\pi$ rad

(c) For x = $\frac{\lambda }{2}$ , we have $\phi =\frac{2\pi }{\lambda }*$ $\frac{\lambda }{2}$ = $\pi$ rad

(d) For x = $\frac{3\lambda }{4}$ , we have $\phi =\frac{2\pi }{\lambda }*$ $\frac{3\lambda }{4}$ = $1.5\pi$ rad

All the waves have different phases. The given transverse harmonic wave is:

y(x, t) = 3.0 sin (36t + 0.018x + $\frac{\pi }{4}$ ) ………(i)

For x = 0, the equation reduces to

y (0,t) = 3.0 sin (36t + $\frac{\pi }{4}$ )

Also $\omega$ = $\frac{2\pi }{T}$ , so T = $\frac{2\pi }{\omega }$ = $\frac{2\pi }{36}$ = $\frac{\pi }{18}$ s

For plotting y vs., t graphs using different values of t, we get

When t = 0, y = 2.12

t = $\frac{T}{8}$ = $\frac{\pi }{18*8}$ , y = 3

t = $\frac{2T}{8}$ = $\frac{2*\pi }{18*8}$ , y = 2.12

t = $\frac{3T}{8}$ = $\frac{3*\pi }{18*8}$ , y = 0

t = $\frac{4T}{8}$ = $\frac{4*\pi }{18*8}$ , y = -2.12

t = $\frac{5T}{8}$ = $\frac{5*\pi }{18*8}$ , y = -3

t = $\frac{6T}{8}$ = $\frac{6*\pi }{18*8}$ , y = -2.12

t = $\frac{7T}{8}$ = $\frac{7*\pi }{18*8}$ , y = 0

For x = 2 and x = 4. The phases of the three waves will get changed, frequency and amplitude will remain unchanged for any change of x.

The equation of a progressive wave travelling from right to left is given by the displacement function:

y (x, t) = a $\sin ?(\omega t+kx+\phi )$ ……….(i)

The given equation is

y(x, t) = 3.0 sin (36 t + 0.018 x + $\pi$ /4) …….(ii)

On comparing equations (i) and (ii), we find that the equation (ii) represents a travelling wave, propagating from right to left. Now, using equations (i) and (ii), we can write

$\omega =36rad/s$ , k = 0.018 $m^{-1}$

From the relation $\nu =\frac{\omega }{2\pi }$ and $\lambda =\frac{2\pi }{k}$ and v = $\nu \lambda$ , we can write

v = $\frac{\omega }{2\pi }*$ $\frac{2\pi }{k}$ = $\frac{\omega }{k}$ = $\frac{36}{0.018}$ = 2000 cm /s = 20 m/s

Hence, the speed of the given travelling wave is 20 m/s

The frequency $\nu =\frac{\omega }{2\pi }$ = $\frac{36}{2\pi }$ = 5.73 Hz

Comparing equation (i) and (ii), we get

$\phi$ = $\frac{\pi }{4}$

The distance between two successive crests or troughs is equal to the wavelength of the wave.

Wavelength $\lambda =\frac{2\pi }{k}$ = $\frac{2\pi }{0.018}$ = 349.07 cm = 3.49 m

In the equation $v=\sqrt{\frac{\gamma P}{\rho }}$ ……(i)

$\rho$ Density = $\frac{Mass}{Volume}$ = $\frac{M}{V}$ where M = molecular weight of the gas, V = Volume of the gas, so we can write

$v=\sqrt{\frac{\gamma PV}{M}}$ …….(ii)

For ideal gas equation, PV = nRT, n = 1 so PV = RT

For constant T, PV = constant

In equation (ii), since PV = constant, $\gamma$ and M constant, v is also constant. Hence, at a constant temperature, the speed of sound in a gaseous medium is independent of the change in the pressure of the gas.

From equation (i) $v=\sqrt{\frac{\gamma P}{\rho }}$

For 1 mole of an ideal gas, the gas equation can be written as PV = RT or P = $\frac{RT}{V}$

Substituting in equation (i), we get $v=\sqrt{\frac{\gamma RT}{\rho V}}$ = $\sqrt{\frac{\gamma RT}{M}}$

Since $\gamma$ , R and M are constant, we get v $\propto \surd T$

Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium. i.e. the speed of sound increase with an increase in the temperature of the gaseous medium and vice versa.

Let $v_m$ and $v_d$ be the speed of sound in moist air and dry air respectively and ${\rho }_m$ and ${\rho }_d$ be the corresponding densities

From equation (i) $v=\sqrt{\frac{\gamma P}{\rho }}$ , we get $v_m\mathrm{}=\sqrt{\frac{\gamma P}{{\rho }_m}}$ and $v_d\mathrm{}=\sqrt{\frac{\gamma P}{{\rho }_d}}$

$\frac{v_m}{v_d}$ = $\sqrt{\frac{{\rho }_d}{{\rho }_m}}$

However, the presence of water vapour reduces the density of air, i.e. ${\rho }_d<$ ${\rho }_m$ , so $v_m<$ $v_d$