What are the values of phase velocity and group velocity respectively of the de- Broglie wave describing a free electron with velocity?
De-Broglie Wave: According to de-Broglie, every moving object is associated with a wave, known as de-Broglie wave. If m is mass of the object moving with velocity v then associated wavelength
λ is a given as
Where h is Planck's constant
The velocity of waveforms of constant phase is called phase velocity and is equal to the velocity of the wave Hence,
phase velocity = λ x frequency ........ (i)
By energy equation mc2 = hv
Frequency = mc2/h....(ii)
By Eqs. (i) and (ii)
Phase velocity = λ * v
Phase velocity = c2/V...(iii)
Now, the group velocity is given as
By differentiation we get,
So, the group velocity is given as,
If the dispersion relation for electromagnetic waves of wave vector k in the ionoshpere is , then the group velocity vg and phase velocity vp are related as :
We have ...(i)
The phase velocity ...(ii)
The group velocity ...(iii)
Differentiate equation, we get
By Eq. (ii) ω= kvp
Eq. (iv) ⇒ ⇒ ⇒
If waves in an ocean travel with a phase velocity then what is the group velocity vg of the ‘wave packet' of these waves?
The phase velocity
where λ = wavelength
From Eqs. (i) and (ii), we have
⇒ Group velocity
The relation between angular frequency 00 and wave number K for given type of waves is . The wave number K0 for which the phase velocity equal the group velocity is
Dispersion relation is
Given vg =vp
A travelling wave is represented by— If die ratio of the wave velocity to the peak particle velocity is 10. then the amplitude A is equal to :
The wave equation is y =
The particle velocity
⇒ The peak velocity of particle = A 2π v ...(i)
and the wave velocity = vλ ...(ii)
The relation between wavelength and frequency for a waveguide is given s calculate the group velocity of the wave .
A person hears the sound of jet aeroplane after it has passed over his head. The angle of the jet plane with the horizontal when the sound appears to be coming vertically downwards is 60°. If the velocity of sound is v, then the velocity of the jet plane should be
Distance covered by the sound to reach the person = distance covered by jet x cot 60°
∴ velocity of jet =
A sound source with frequency f0 moves with a speed u towards a stationary wall. The speed of sound is v. The frequency detected by an observer moving together with source with same speed u is
The frequency of sound received by the wall is given by
Now, the wall acts as a source of sound. The apparent frequency detected by the observe is
A source of sound approaches an observer and then recedes from it. Ratio of frequencies of sound as the source approaches and as the source recedes is 6 : 5. Find the velocity of source (Velocity of sound = 330 ms-1)
Let source of sound moves with speed vs and its real frequent When source approaches observer frequency is given as
v → velocity of sound
⇒ v = 330
When source recedes
By Eqs. (i) and (ii) we get
⇒ vs = 30 m/s
A fire alarm sounds with a frequency of 480 Hz. Two fire engines dash to the site to extinguish the fire from opposite directions. One travels with a speed of 33 m/s and the other with 27m/s. If the velocity of sound in air be 330 m/s, the difference between the frequencies of the sirens are heard by the drivers of the two fire engines will be :
A particle is executing simple harmonic motion. What is the nature of the graph of velocity as a function of displacement?
equation of an ellipse.
Consider the following statements about a harmonic oscillator: -
1. The minimum energy of the oscillator is zero.
2. The probability of finding it is maximum at the mean position.
Which of the statement given above is/are correct ?
We know that total energy
What is the phase difference between the two simple harmonic vibrations represent by .
A particle is executing SHM. If the displacement at any instant is given by x = 3 c o s 2 + 4 s in 2 t. What is the time period of the particle?
x = 3 cos 2t + 4 sin 2t
x = 3 cos 2t + 4 sin 2t
⇒ ω = 2
An object of small size and mass m is attached to a spring of force constant K fixed at one end and is undergoing linear oscillatory motion. If its maximum displacement from the point of equilibrium is x. What is its speed, when it is at A = x0
A = x0
Consider two waves represented by two mutually perpendicular electric field vectors : Their superposition will result in a plane polarized light, if:
The ordinary light is a wave in which its both component E and B vibrate perpendicularly to the direction of the light. If E vibrate in a plane the light is said to plane polarised light.
The two perpendicular vibrating lightwaves form a plane polarised light if their frequencies are equal and they have a phase difference of π/2.
In the arrangement shown in figure, the particle m1 rotates in a radius r on a smooth horizontal surface with angular velocity ω0. Then choose the correct statement.
(i) Considering the equilibrium of m2, we have
(ii) Let the mass m2 is displaced downwards by a distance x. Now the radius of the circular path decreases by x and the angular speed of m1 increases. Applying the conservation of angular momentum, we have
The tension is also increased as shown below :
As a result, m2 gets a restoring force, given by
F = - [T - m2g]
According to second law of motion. F is given by
From eq. (4) and (5). we have
Substituting the value of ω0 from eq. (1). we get
This represents S.H.M. The time period T is given by
A solid cylinder of mass M is attached to horizontal spring of force constant k as shown in figure. The spring is first stretched by.
length l, by moving cylinder to the left. When the system is released from rest, the cylinder rolls along the horizontal surface without slipping. Centre of mass of the cylinder executes simple harmonic motion.
Let X0 be the equilibrium length of the spring. This is stretched by a length L So. the initial length of the spring xi = (X0 + l).
Potential energy stored in spring
When the spring is released, say att = 0. the cylinder rolls without slipping being pulled by the spring. Consider the situation at time t. Letthe centre of mass be at a distance x from the oiigin O. At this instant.
K .E . (translational) of cylinder = 1/2 mv2
and K.E. (rotational) of cylinder
where v = linear velocity of cylinder
∴ Total energy of cylinder
Instantaneous extension of spring = (x - x0)
Potential energy of the spring
∴ Loss of P.E. =
Applying the law of conservation of energy, we have gain in K.E. = loss is potential energy
Differentiating on both sides with respect to t and considering l and x0 constant we get
As x is constant this equation can be written as
This is equation of S.H.M.
At equilibrium position, the potential energy is completely converted to kinetic energy
A progressive and a stationary simple harmonic wave each has the same frequency of 250Hz, and the same velocity of 30 m/s. Then which of the following are correct.
Given, n = 250 Hz, v = 30 m/s
(a) Phase difference between two points at a distance λ = 2π
Phase difference between two points, unit distance apart
∴ Phase difference for a distance of 10 cm
The general equation of a plane progressive wave is given by
Here y = 0.03 sin 2π (250 t - 25x/3)
(c) The distance between nodes in stationary wave
Equation of stationary wave is given by
∴ y = 0.02
The equation for the displacement of a stretched string is given by y = 4 sin where y and x are in cm and t is in sec . Which of the following are correct.
Comparing the given equation with the general wave equation:-
we find that: -
(A) As there is negative sign between t and x terms, the wave is propagating along positive x-axis.
(B) The amplitude of the wave A = 4 cm = 0.04 m.
(C) The time period of the wave T = 0.02 s = (1/50)s.
(D) The frequency of the wave f = (1/T) = 50Hz.
(e) Angular frequency of the wave
(f) The wavelength of the wave
(g) The propagation const. (= wave vector =
(h) The velocity of wave
(i) The phase const., i.e., initial phase
(j) The max. particle velocity
The spherical surface of a plane -convex lens of radius of curvature R = 2m is gently placed on a flat plate. The space between them is filled with a transparent liquid of refractive index 1.5. The refractive indices of the lens and the flat are 1.4 and 1.6 respectively. The radius of the fifteen dark newton's ring in the reflected light of wavelength λ is found to be √5m.m. Determine the wave length λ (in microns) of the light.
We know for minima in newtons ring
where n=1. 2. 3
(a) for n = 15
e = thickness of liquid film = x2/2R
X = radius of ring
R = radius of curvature of lens one gets
Given μ = 1.5
R = 2m
n = 15
So. λ = 0.25 * 10-6 m
The ratio in the densities of oxygen and nitrogen is 16 : 14. At what temperature (in °C) the speed of sound will be the same which is in nitrogen at 15°C.
If M is the molecular weight of the gas and T is the absolute temperature, then speed of sound in a gas.
where R is the universal gas constant.
Velocity of sound in oxygen at t°C
Velocity of sound in nitrogen at 15°C
According to the given problem
Solving we get t = 56.1°C.
The speed of sound at NTP in air is 332 m/sec. Calculate the speed (in in/ s) of sound in hydrogen at NTP (Air is 16 times heavier than hydrogen.)
We know that the speed of sound in a gas is given by
If va and vh be the velocities of air and hydrogen respectively at normal pressure and temperature, then
where dh and da are densities of hydrogen and air respectively.
The speed of sound at NTP in air is 332 m/sec. Calculate the speed (in m/ s) of sound in hydrogen at 8190C temperature and 4 atmospheric pressure (Air is 16 times heavier than hydrogen.)
We know that the speed of sound is directly proportional to the square root of the absolute temperature of the gas. Let v0 and v819 be the speeds of sound in hydrogen at 0°C and 819°C respectively, then
∴ v819 = 2 * v0 = 2 * 1328 = 2656 m/sec
The power (in diopters) of an equiconvex lens with radii of curvature of 10 cm and refractive index o f 1.6 i s ________ .
The focal length of a lens is given as
Here, n = 1.6. R1 =0.1 m. R2 = - 0.10 m
= (0.6) (20)
⇒ Power P = 1/f = +12D
A parallel beam of light of diameter 1.8cm contains two wavelengths 4999.75 and 5000.25 . The light is incident perpendicularly on a large diffraction grating with 5000 lines per centimeter using Rayleigh criterion the least order at which the two wave / lengths are resolved i s _______ .
So resolving power of grating
R = Nn ≥ 10000
n ≥ 1.11
n = 2
A diffraction grating of length 2.5 - 1 0-2 m is illuminated by a light with two wavelengths 5997 and 6003 . The maximum size of the grating element d ( in μm ) required to resolve the two wavelengths in the first order is ______ .
Resolving power required =λ/μ
= 6000/6 = 1000
R = Nn = 1000
R = Nn = 1000
⇒ N = 1000
A plane monochromatic light wave falls normally on a diaphrapm with two narrow slits separated by a distance d = 2.5 mm. A fringe pattern is formed on a screen placed at a distance I = 100cm behind the diaphrapm. By what distance and in which direction will these fringes be displaced when one of the slits is covered by a glass plate of thickness h = 10μm (in mm).
D = 10 cm = 1 m
d = 2.5 mm = 2.5 x 10-3 m
t = 10 x 10-6 m , n = 1.5
A parallel beam of EM waves consistings of two wavelength 14000 and 26000 coherent in themselves falls on a double slit apparatus. The separation between the two slits is 2 cm and that between plane of the slits at screen is 1 metre.
Find out the separation between the 2nd maxima formed by each wavelengths, (in 10-4 m)
The figure shows a YDSE arrangment. Given :d = 1 cm. D = 100cm, λ = 5000. If the screen is given an instantaneous velocity of 1 mm per second towards right from the slits.
Then what is the rate of change of fringe width?
in (μ cm/sec).