1. The displacement in a string is given by the following
equation:...where a, λ and ν represent the amplitude, wavelength and the
frequency of the wave. Assume a = 0.1 cm, λ = 4 cm, ν = 1 sec –1. Plot
the time dependence of the displacement at x = 0, 0.5 cm, 1.0 cm, 1.5
cm, 2 cm, 3 cm and 4 cm. Interpret the plots physically.
Get solution 2.
The displacement associated with a standing wave on a sonometer is
given by the following equation:...If the length of the string is L then
the allowed values of λ are 2L, 2L/2, 2L/3, … (see Sec. 13.2). Consider
the case when λ = 2L/5; study the time variation of displacement in
each loop and show that alternate loops vibrate in phase (with different
points in a loop having different amplitudes) and adjacent loops
vibrate out of phase.
Get solution 3.
A tunnel is dug through the earth as shown in Fig. 7.15. A mass is
dropped at the point A along the tunnel. Show that it will execute
simple harmonic motion. What will the time period be?...
Get solution 4.
A 1 g mass is suspended from a vertical spring. It executes simple
harmonic motion with period 0.1 sec. By how much distance had the spring
stretched when the mass was attached?
Get solution 5.
A stretched string is given simultaneous displacement in the x- and y-
directions such that...Study the resultant displacement (at a particular
value of z) as a function of time.
Get solution 6. In the above problem, if...what will be the resultant displacement?
Get solution 7.
As mentioned in Sec.7.5, alkali metals are transparent to ultraviolet
light. Assuming that the refractive index is primarily due to the free
electrons and that there is one free electron per atom, calculate ...for
Li, K and Rb; you may assume that the atomic weights of Li, K and Rb
are 6.94, 39.10 and 85.48 respectively; the corresponding densities are
0.534, 0.870 and 1.532 g/cm3. Also, the values of various physical
constants are: m = 9.109 × 10 – 31 kg, q = 1.602 × 10 – 19 C and ε0 =
8.854 × 10 – 12 C/N-m2.[Ans: 1550 Å, 2884 Å and 3214 Å; the
corresponding experimentalvalues are 1551 Å, 3150 Å and 3400 Å
respectively].
Get solution 8.
(a) In a metal, the electrons can be assumed to be essentially free.
Show that the drift velocity of the electron satisfies the following
equation... where ν represents the collision frequency. Calculate the
steady state current density (J = – N q v) and show that the
conductivity is given by ...(b) If r represents the displacement of
the electron, show that ...which represents the polarization. Using
the above equation show that ...which represents the dielectric
constant variation for a free electron gas.
Get solution 9.
Assuming that each atom of copper contributes one free electron and
that the low frequency conductivity σ is about 6 ×107 mhos/meter, show
that ν ≈ 4 × 10 13 s –1. Using this value of ν, show that the
conductivity is almost real for ω 11 s –1. For ω = 10 8 s–1 calculate
the complex dielectric constant and compare its value with the one
obtained for infra-red frequencies.It may be noted that for small
frequencies, only one of the electrons of a copper atom can be
considered to be free. On the other hand, for X-ray frequencies all the
electrons may be assumed to be free (see Problems 7.10, 7.11 and 7.12).
Discuss the validity of the above argument.
Get solution 10.
Show that for high frequencies (ω ≫ ν) the dielectric constant (as
derived in Problem 7.8) is essentially real with frequency dependence of
the form...where ...is known as the plasma frequency. The above
dielectric constant variation is indeed valid for X-ray wavelengths in
many metals. Assuming that at such frequencies all the electrons an be
assumed to be free, calculate ωp for copper for which the atomic number
is 29, mass number is 63 and density is 9 g/cm3. [Ans: ~ 9 × 10
16 sec – 1]
Get solution 11.
Obtain an approximate value for the refractive index of metallic sodium
corresponding to λ= 1 Å. Assume all the electrons of sodium to be free.
Get solution 12.
In an ionic crystal (like NaCl, CaF2, etc.) one has to take into
account infra-red resonance oscillations of the ions. Show that Eq. (68)
modifies to...where M represents the reduced mass of the two ions and p
represents the valency of the ion (p = 1 for Na+, Cl –; p = 2 for Ca++,
F2– –). Show that the above equation can be written in the
form* ...where ... ...
Get solution 13.
The refractive index variation for CaF2 (in the visible region of the
spectrum) can be written in the form...where λ is in meters.(a) Plot the
variation of n2 with λ in the visible region.(b) From the values of A1
and A2 show that m/M ≈ 2.07 × 10– 5 and compare this with the exact
value.(c) Show that the value of n∞ obtained by using the constants A1,
A2, λ1 and λ2 agrees reasonably well with the experimental value.
Get solution 14.
(a) The refractive index of a plasma (neglecting collisions) is
approximately given by (see Sec. 7.6)...where ...s –1is known as the
plasma frequency. In the ionosphere the maximum value of N0 is ≈ 10 10 –
10 12 electrons/m3. Calculate the plasma frequency. Notice that at high
frequencies n2 ≈ 1; thus high frequency waves (like the one used in TV)
are not reflected by the ionosphere. On the other hand, for low
frequencies, the refractive index is imaginary (like in a conductor –
see Sec. 24.3) and the beam gets reflected. This fact is used in long
distance radio communications (see also Fig. 3.27).(b) Assume that for x
≈ 200 km, N = 10 12 electrons/m3 and that the electron density
increases to 2 × 10 12 electrons/m3 at x ≈ 300 km. For x N, plot the
corresponding refractive index variation.
Get solution 