Problems on properties of waves - class-XII

The phase difference between the vibrations of two particles of the medium is given by :

Maximum amplitude A_max = A1 + A2 = (x + 1) * A_0. Minimum amplitude A_min = |A1 - A2| = (x - 1) * A_0. The difference is A_max - A_min = (x + 1) * A_0 - (x - 1) * A_0 = 2 * A_0.

For the given wave, $k=0.1\pi=\dfrac{2\pi}{\lambda}$
$\implies \lambda=20m$

The phase difference delta phi is given by (2pi / lambda) * delta x, or from the wave equation k * delta x. Here k = 0.01 pi, so delta phi = 0.01 pi * 25 = 0.25 pi = pi / 4 radians.

${y _1} = 2\sin \left( {\omega t - kx} \right)$
${y _2} = 2\cos \left( {\omega t - kx} \right) = 2\sin \left( {\omega t - kx + \frac{\pi }{2}} \right)$
So, Phase difference = $\frac{\pi}{2}$
Hence sources are incoherent.

Phase difference $= \dfrac{2\pi}{\lambda}\times$path difference. 
The phase difference between any two particles in a wave determines lack of harmony in the vibrating state of two particles ie, how far one particle leads the other or lags behind the other. 
From relation 
$\Delta \phi =\dfrac{2\pi}{\lambda}\times \delta x$

For a transverse wave, a phase change of $\pi$ occurs when it is reflected from a denser medium.

$\omega =\dfrac{2\pi}{T}$
$T=\dfrac{2\pi}{\omega}$
$T=1sec$
$\therefore 2\pi \ rad \ in \   1 sec$
$in \ 0.4 sec \  2\pi\times 0.9$
$=144^o$

phase difference between two successive compression is $2\pi $
$\therefore $ phase difference between a particle at one compression and in third compression is $2(2\pi)= 4\pi $

Reflections of light at the interface between media often produce phase differences. The phase difference between incident and reflected wave is $2\pi$ and $\pi / 2$ radians.

Phase difference between two successive compression of rarefaction is $2\pi $
As  rarefaction appears between two compression, phase difference is $\pi $.

Resultant amplitude A_res = sqrt(A1^2 + A2^2 + 2 * A1 * A2 * cos(phi)). For A1 = A2 = A, A_res = 2 * A * cos(phi / 2). For A_res = 2 * A, cos(phi / 2) = 1, so phi / 2 = 0, which means phi = 0.

Phase difference between a crest and its successive trough is $ \pi $radians

The frequency of the wave is $f = v/\lambda=400/0.5=800 Hz$

Time period = 1/800 = 1.25 ms

Thus, both the points are one time period apart. Hence their phase difference will be zero or $2 \pi$

The correct option is (a)

Beats period$=\cfrac{1}{30-20}=0.1second\ \Delta Y=\cfrac{2\pi}{T}\Delta t\ \quad=\cfrac{2\pi}{0.1}\times0.6=2\pi\times6=12\pi\ \therefore\Delta Y=12\pi$

Phase difference = $(2\pi/5)$ path difference $\implies$ phase difference = $6 \pi/5$

The correct option is (b) 

Given,

Given,

At t=1, the phase of Y1 is pi/2 + phi. The phase of Y2 is 2pi/3 + phi. The difference is 2pi/3 - pi/2 = 4pi/6 - 3pi/6 = pi/6.

${X _1} = A\sin \left( {wt + {\phi _1}} \right)$

Given,

Two consecutive crest =$5\,m$.

Two wave cross in time, $t=\,1\sec $

Wavelength = distance between two consecutive crest

$\lambda =5\,m$

Velocity, $v=\dfrac{2\lambda }{t}=\dfrac{2\times 5}{1}=10\,m{{s}^{-1}}$

Hence, wave speed is $10\,m{{s}^{-1}}$

A 60° phase difference represents 60/360 = 1/6 of a complete cycle. For a 60 Hz AC supply, the time period T = 1/f = 1/60 seconds. The time difference for 60° = (1/6) × (1/60) = 1/360 seconds. Option D matches this calculation.

Phase difference is related to path difference by delta phi = (2pi / lambda) * delta x. Given delta phi = 0.5 pi and delta x = 0.8 m, we find lambda = 3.2 m. Wave velocity v = frequency * wavelength = 120 * 3.2 = 384 ms^-1.

Convert y2 to sine form: y2 = a sin(omegat + kx + pi/2). The phase of y1 is omegat + kx + 0.57. The phase of y2 is omega*t + kx + 1.57. The difference is 1.57 - 0.57 = 1.0 radian.

The amplitude of the resultant of two waves with equal amplitudes a and phase difference phi is given by A = sqrt(2a^2 + 2a^2 cos(phi)). Setting the resultant amplitude equal to a gives 1 + cos(phi) = -1/2, leading to cos(phi) = -1/2, so phi = 2pi / 3.

At x = A/2, the phase angles are pi/6 and 5pi/6. Since they move in opposite directions, one is at pi/6 (moving away from equilibrium) and the other is at 5pi/6 (moving toward equilibrium). The difference is 5pi/6 - pi/6 = 4pi/6 = 2pi/3.

For the same phase, the path difference must be a multiple of the wavelength. Here, k = 70, so 2pi/lambda = 70, lambda = 2pi/70 m = 2pi/70 * 100 cm = 20pi/7 cm.

Rewriting x = theta * sin(alpha + pi/6) and x' = theta * cos(alpha) = theta * sin(alpha + pi/2), the phase difference between the two sine waves is the difference in their phase angles, which is (alpha + pi/2) - (alpha + pi/6) = pi/3.

v1 = dy1/dt = 0.1 * 100pi * cos(100pit + pi/3). v2 = dy2/dt = -0.1 * pi * sin(pit). This question is garbled regarding the frequencies (100pi vs pi), making a standard phase difference calculation between them invalid.

A progressive wave must be a function of (vt - x) or (t - x/v). Option D uses (t/T - x/v), which is dimensionally inconsistent if T is period and v is velocity, as t/T is dimensionless but x/v is time.

y = 0.1 sin(60t + 2x). omega = 60, k = 2. Frequency f = omega/2pi = 60/2pi = 30/pi Hz. Wavelength lambda = 2pi/k = 2pi/2 = pi m. Amplitude = 0.1 m = 10 cm. The wave moves in negative x direction because of the + sign.

Particle velocity v_p = dy/dt = y0 * omega * cos(omega*t - kx). Max particle velocity = y0 * omega. Wave velocity v_w = omega/k. Given y0 * omega = 2 * (omega/k), so y0 = 2/k. Since k = 2pi/lambda, y0 = 2 / (2pi/lambda) = lambda/pi. Thus, lambda = pi * y0.

If one boat is at crest and other at trough, the distance 10m is half the wavelength (lambda/2). So lambda = 20m. Period T = 4s. Speed v = lambda/T = 20/4 = 5 m/s.

In non-inertial frames, the relation L = I*omega is generally not valid due to the changing nature of the moment of inertia or frame-dependent definitions of angular momentum. However, the torque equation dL/dt = Gamma is a fundamental law of motion that holds in inertial frames, but in non-inertial frames, pseudo-torques must be included.

Using sin^2(theta) = (1 - cos(2theta))/2, the equation becomes y = a/2 - (a/2)cos(4pint - 4pix/lambda). This represents a wave with amplitude a/2, frequency 2n, and wavelength lambda/2 (since k = 4pi/lambda = 2pi/lambda_new).

For a string rotating in a circle, the wave speed relative to the string is v0. In the absence of gravity, the speed of a transverse wave on a rotating hoop is equal to the tangential speed v0.

Using cos^2(theta) = (1 + cos(2theta))/2, the equation becomes y = A/2 + (A/2)cos(4pint - 4pix/lambda). This corresponds to amplitude A/2, frequency 2n, and wavelength lambda/2.

In the argument omega(x/v - k), the dimensions of omega are T^-1. For the argument to be dimensionless, (x/v - k) must have dimensions of T. Since x/v is distance/velocity = time, k must also have dimensions of time.

The wave equation is y = a sin(200t - x). The standard form is y = a sin(omegat - kx). Here omega = 200 and k = 1. Wave velocity v = omega/k = 200/1 = 200 m/s.

The relative deformation (strain) is dy/dx. y = 4 sin(2pit/0.02 - 2pix/100). dy/dx = 4 * (-2pi/100) * cos(...) = -0.08pi * cos(...). The amplitude of this is 0.08pi.

In a wave, a full cycle consists of a crest and a trough, spanning one wavelength lambda. The distance from a crest to an adjacent trough represents half of a wavelength, which is 3 / 2 = 1.5 m.

Wavelength of the wave=$\lambda=\dfrac{v}{\nu}=\dfrac{300}{25}=12m$

y1 = 10 sin(3pit + pi/4), amplitude = 10. y2 = 5(3 sin(3pit) + sqrt(3) cos(3pi*t)). Using R = sqrt(A^2 + B^2 + 2AB cos(phi)), y2 = 5 * sqrt(3^2 + sqrt(3)^2) * sin(...) = 5 * sqrt(9+3) = 5 * sqrt(12) = 10 * sqrt(3). Ratio = 10 / (10 * sqrt(3)) = 1/sqrt(3).

Equation for a travelling harmonic wave is given as:

The wavelength of the wave can be found by using $\dfrac{\lambda}{T}=v$

The general equation for a wave moving in the positive x-direction is y = A sin(2pi(ft - x/lambda)). Given A = 0.2, f = velocity/lambda = 360/60 = 6 Hz, and lambda = 60, the equation becomes y = 0.2 sin(2pi(6t - x/60)).

Resultant intensity for two interfering waves of intensity I_0 and phase difference phi is I = I_1 + I_2 + 2 * sqrt(I_1 * I_2) * cos(phi). With I_1 = I_2 = I_0 and phi = 60 degrees, cos(60) = 1/2, yielding I = I_0 + I_0 + I_0 = 3 I_0.

Here, we have given: $y=3sin\omega t+4cos\omega t$

Phase difference between two points in a standing wave =$n\pi$