physics ncert solutions class 11th

(a) The general equation representing a stationary wave is given by the displacement function y(x,t) = 2asin kx cos $\omega t$

The equation is similar to the give equation y(x, t) = 0.06 sin ( $\frac{2\pi }{3}x)$ cos (120 $\pi t)$

Hence, the given function represents a stationary wave.

A wave travelling along the positive x-direction is given as: $y_1$ = asin ( $\omega$ t – kx)

The wave travelling along the negative x-direction is given as: $y_2$ = asin ( $\omega$ t + kx)

(b) The superposition of these two waves yields:

y = $y_1+y_2$ = asin ( $\omega$ t – kx) + asin ( $\omega$ t + kx)

= a sin( $\omega$ t)cos(kx) – a sin(kx)cos( $\omega$ t) – asin( $\omega$ t)cos(kx) – asin(kx)cos( $\omega$ t)

= -2a sin (kx) cos ( $\omega$ t)

= -2asin( $\frac{2\pi }{\lambda }$ x) cos (2 $\pi \nu$ t) …….(i)

The transverse displacement of the string is given as

y(x, t) = 0.06 sin ( $\frac{2\pi }{3}x)$ cos (120 $\pi t)\dots \dots \dots$ (ii)

Comparing equations (i) and (ii), we have $\frac{2\pi }{\lambda }$ = $\frac{2\pi }{3}$

Therefore $\lambda$ = 3m, it is given that 120 $\pi$ = 2 $\pi \nu$ so Frequency, $\nu =60\mathrm{H}\mathrm{z}$

Wave speed, v = $\nu \lambda$ = 60 $*3$ = 180 m/s

The velocity of a transverse wave travelling in a string is given by the relation:

v = $\sqrt{\frac{T}{\mu }}$ ….(i)

We have v = 180 m/s, mass of the string, m = 3 $*{10}^{-2}$ kg,

length of the string, l = 1.5 m

Mass per unit length of the string, $\mu$ = m/l = $\frac{3}{1.5}*{10}^{-2}$ = $2*{10}^{-2}$ kg/m

(c) Tension in the string, T = $v^2$ $*\mu ={180}^2$ $*2*{10}^{-2}$ = 648 N

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