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#### Page No 321:

No, in wave motion there is no actual transfer of matter but transfer of energy between the points where as when wind blows air particles moves with it.

#### Page No 321:

It is a non-mechanical wave because this type of wave does not require a material medium to travel.

#### Page No 321:

Equation of the wave is

When the variable of the equation is (c2x + c3t), then the wave must be moving in the negative x-axis with time t.

#### Page No 321:

Equation of the wave is given by
$y=A\mathrm{sin}\left(\omega t-kx\right)$
where
A is the amplitude
ω is the angular frequency
k is the wave number
Velocity of wave, $v=\frac{\omega }{k}$
Velocity of particle,
Max velocity of particle, ${v}_{{p}_{\mathrm{max}}}=A\omega$
As given
$A<\frac{\lambda }{2\mathrm{\pi }}$

#### Page No 321:

When two wave pulses identical in shape but inverted with respect to each other meet at any instant, they form a destructive interference. The complete energy of the system at that instant is stored in the form of potential energy within it. After passing each other, both the pulses regain their original shape.

#### Page No 321:

To prove,

$\frac{{y}_{\mathrm{max}}}{{v}_{\mathrm{max}}}=\frac{{v}_{\mathrm{max}}}{{a}_{\mathrm{max}}}\phantom{\rule{0ex}{0ex}}\mathrm{LHS}\phantom{\rule{0ex}{0ex}}\frac{{y}_{\mathrm{max}}}{{v}_{\mathrm{max}}}=\frac{A}{A\omega }=\frac{1}{\omega }\phantom{\rule{0ex}{0ex}}\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\frac{{v}_{\mathrm{max}}}{{a}_{\mathrm{max}}}=\frac{A\omega }{{\omega }^{2}A}=\frac{1}{\omega }$

No, componendo and dividendo is not applicable. We cannot add quantities of different dimensions.

#### Page No 321:

Equation of the wave: y = sin(kxωt + Φ)
Here, A is the amplitude, k is the wave number, ω is the angular frequency and Φ is the initial phase.

The argument of the sine is a phase, so the smallest positive phase constant should be

Therefore, the smallest positive phase constant is 1.5π.

#### Page No 321:

Yes, at the centre. The centre position is a node. If the string vibrates in its first overtone, then there will be two positions, i.e., two nodes, one at x = 0 and the other at x = L.

#### Page No 322:

(c) $\lambda /2$ A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. The speeds at the maximum point and at the minimum point are same although the direction of motion are different. The difference between the positions of maxima and minima is equal to $\lambda /2$.

#### Page No 322:

(c) $\lambda /2$

A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. Therefore, applying similar argument we can say that if a particular particle has zero displacement at a certain instant, then the particle closest to it having zero displacement is at a distance is equal to $\lambda /2$.

#### Page No 322:

(a)

Here x is the particle displacement of the wave and the wave is travelling along the Y-axis because the particle displacement is perpendicular to the direction of wave motion.

#### Page No 322:

(b) amplitude A/2, frequency $\omega /\pi$

$y=A{\mathrm{sin}}^{2}\left(kx-\omega t\right)$

$\left[{\mathrm{cos}}^{2}\theta =1-2{\mathrm{sin}}^{2}\theta \phantom{\rule{0ex}{0ex}}{\mathrm{sin}}^{2}\theta =\frac{1-{\mathrm{cos}}^{2}\theta }{2}\right]$

$y=A\left[\frac{1-{\mathrm{cos}}^{2}\left(kx-\omega t\right)}{2}\right]\phantom{\rule{0ex}{0ex}}y=\frac{A}{2}\left[1-{\mathrm{cos}}^{2}\left(kx-\omega t\right)\right]$
Thus, we have:
Amplitude = $\frac{A}{2}$
Frequency = $2\left(\frac{\omega }{2\pi }\right)=\frac{\omega }{\pi }$

#### Page No 322:

(d) Sound waves

There are mainly two types of waves: first is electromagnetic wave, which does not require any medium to travel, and the second is the mechanical wave, which requires a medium to travel. Sound requires medium to travel, hence it is a mechanical wave.

#### Page No 322:

(a) $\nu$

The boat transmits the same wave without any change of frequency to cause the cork to execute SHM with same frequency though amplitude may differ.

#### Page No 322:

(a) 1/2

Wave speed is given by
$\nu =\sqrt{\frac{T}{\mathit{µ}}}$
where
​T is the tension in the string
v is the speed of the wave
μ is the mass per unit length of the string
$µ=\frac{M}{L}=\rho \frac{V}{L}=\rho \frac{\left(AL\right)}{L}$
where
M is the mass of the string, which can be written as ρV
L is the length of the string
$=\rho \left(\pi {r}^{2}\right)=\rho \left(\pi \frac{{D}^{2}}{4}\right)\phantom{\rule{0ex}{0ex}}\therefore \nu =\sqrt{\frac{T}{\rho \pi \frac{{D}^{2}}{4}}}=\frac{2}{D}\sqrt{\frac{T}{\rho \pi }}$
where D is the diameter of the string.
Thus, v $\frac{1}{D}$
Since, rA = 2rB

From Equations (1) and (2) we get
$\frac{{v}_{\mathrm{A}}}{{v}_{\mathrm{B}}}=\frac{1}{2}$

#### Page No 322:

(d) $1/\sqrt{2}$ ${T}_{\mathrm{AB}}=T\phantom{\rule{0ex}{0ex}}{T}_{\mathrm{CD}}=2T$
where
​TAB is the tension in the string AB
​TCD is the tension in the string CD
The eelation between tension and the wave speed is given by
$v=\sqrt{\frac{T}{\mathit{µ}}}\phantom{\rule{0ex}{0ex}}v\propto \sqrt{T}$
where
v is the wave speed of the transverse wave
μ is the mass per unit length of the string
$\frac{{v}_{1}}{{v}_{2}}=\sqrt{\frac{T}{2T}}=\frac{1}{\sqrt{2}}$

#### Page No 322:

(d) meaningless

Sound wave is a mechanical wave; this means that it needs a medium to travel. Thus, its velocity in vacuum is meaningless.

#### Page No 322:

(c) $\lambda \text{'}<\lambda$ As $v=\sqrt{\frac{f}{\mathit{µ}}}$
A wave pulse travels faster in a thinner string.
The wavelength of the transmitted wave is equal to the wavelength of the incident wave because the frequency remains constant.

#### Page No 322:

(b) $\sqrt{2}a$

We know that the resultant of the amplitude is given by

For the particular case, we can write
$=\sqrt{{a}^{2}+{a}^{2}+2{a}^{2}\mathrm{cos}\frac{\mathrm{\pi }}{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{2}a$

#### Page No 322:

(d) the information is insufficient to find the relation between t1 and t2.

$v=\sqrt{\frac{\eta }{\rho }}$
But because the length of wires A and B is not known, the relation between A and B cannot be determined.

#### Page No 322:

(b) the velocity but not for the kinetic energy

The principle of superposition is valid only for vector quantities. Velocity is a vector quantity, but kinetic energy is a scalar quantity.

#### Page No 322:

(d) The pulses will pass through each other without any change in their shapes.

The pulses continue to retain their identity after they meet, but the moment they meet their wave profile differs from the individual pulse.

#### Page No 322:

(b) 2A2

We know resultant amplitude is given by
${A}_{\mathrm{net}}=\sqrt{{A}_{1}^{2}+{A}_{2}^{2}+2{A}_{1}{A}_{2}\mathrm{cos}\varphi }$
For maximum resultant amplitude
${A}_{\mathrm{max}}={A}_{1}+{A}_{2}$
For minimum resultant amplitude
${A}_{\mathrm{min}}={A}_{1}-{A}_{2}$

So, the difference between Amax and Amin is
${A}_{\mathrm{max}}-{A}_{\mathrm{min}}={A}_{1}+{A}_{2}-{A}_{1}+{A}_{2}=2{A}_{2}$

#### Page No 322:

(d) between 0 and 2A

The amplitude of the resultant wave depends on the way two waves superimpose, i.e., the phase angle (φ). So, the resultant amplitude lies between the maximum resultant amplitude (Amax) and the minimum resultant amplitude (Amin).
Amax = A + A = 2A
Amin = AA = 0

#### Page No 322:

(a) A

We know the resultant amplitude is given by

#### Page No 322:

(a) inverse of its length

The relation between wave speed and the length of the string is given by
$v=\frac{1}{2l}\sqrt{\frac{F}{\mathit{µ}}}\phantom{\rule{0ex}{0ex}}$
where
l is the length of the string
F is the tension
μ linear mass density
From the above relation, we can say that the fundamental frequency of a string is proportional to the inverse of the length of the string.
$v\propto \frac{1}{l}$

#### Page No 323: Given,
Speed of the wave pulse passing on a string in the negative x-direction = 40 cms−1
As the speed of the wave is constant, the location of the maximum after 5 s will be
s = v × t
= 40 × 5
= 200 cm (along the negative x-axis)
Therefore, the required maximum will be located after x = −2 m.

#### Page No 323:

Given,
Equation of the wave travelling on a string stretched along the X-axis:

(a) The dimensions of A (amplitude), T (time period) and $a=\frac{\lambda }{2\pi }$, which will have the dimensions of the wavelength, are as follows:

(b) Wave speed,

(c) If , then the wave travels in the negative direction; and if , then the wave travels in the positive direction.
Thus, we have:

Hence, the wave is travelling is the negative direction.

(d) Wave speed, $v=\frac{a}{t}$
Maximum pulse at t = T  =
Maximum pulse at t = 2T =
Therefore, the wave is travelling in the negative x-direction.

#### Page No 323:

Given,
Wave pulse at t = 0 Wave speed = 10 cms−1
Using the formula $s=v×t$, we get:

#### Page No 323:

(b) 480 Hz

The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be natural frequency of the wire, then standing waves with large amplitude are set up in it.

#### Page No 323:

(b) 480 Hz

The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be the natural frequency of the wire, standing waves with large amplitude are set in it.

#### Page No 323:

(b) vibrate with a frequency of 208 Hz

According to the relation of the fundamental frequency of a string
$\nu =\frac{1}{2l}\sqrt{\frac{F}{\mu }}$
where
l is the length of the string
F is the tension
μ is the linear mass density

We know that ν1 = 416 Hz, l1 = l and l2 = 2l.

#### Page No 323:

(d) 16 kg

According to the relation of the fundamental frequency of a string
$\nu =\frac{1}{2l}\sqrt{\frac{F}{\mu }}$
where l is the length of the string
F is the tension
μ is the linear mass density of the string
We know that ν1 = 416 Hz, l1 = l and l2 = 2l.
Also, m1 = 4 kg and m2 = ?

So, in order to maintain the same fundamental mode
${\nu }_{1}={\nu }_{2}$
squaring both sides of equations (1) and (2) and then equating

#### Page No 323:

(c) may move on the X-axis
(d) may move on the Y-axis

A mechanical wave is of two types: longitudinal and transverse. So, a particle of a mechanical wave may move perpendicular or along the direction of motion of the wave.

#### Page No 323:

(d) in the XY plane

In a transverse wave, particles move perpendicular to the direction of motion of the wave. In other words, if a wave moves along the Z-axis, the particles will move in the XY plane.

#### Page No 323:

(d) be polarised

A longitudinal wave has particle displacement along its direction of motion; thus, it cannot be polarised.

#### Page No 323:

(b) may be longitudinal
(d) may be transverse

Particles in a solid are very close to each other; thus, both longitudinal and transverse waves can travel through it.

#### Page No 323:

(a) must be longitudinal

Because particles in a gas are far apart, only longitudinal wave can travel through it.

#### Page No 323:

(b) A and B move in opposite directions.
(d) The displacements at A and B have equal magnitudes. A and B have a phase difference of π. So, when a sine wave passes through the region, they move in opposite directions and have equal displacement. They may be separated by any odd multiple of their wavelength.
$\stackrel{\to }{{y}_{A}}=A\mathrm{sin}\left(\omega t\right)$
$\stackrel{\to }{{y}_{B}}=B\mathrm{sin}\left(\omega t+\mathrm{\pi }\right)$

#### Page No 323:

(c) Frequency = 25/π Hz
(d) Amplitude = 0⋅001 mm

Equating the above equation with the general equation, we get:
$y=A\mathrm{sin}\left(\omega t-kx\right)\phantom{\rule{0ex}{0ex}}\omega =\frac{2\mathrm{\pi }}{T}=2\pi v\phantom{\rule{0ex}{0ex}}k=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}$
Here, A is the amplitude, ω is the angular frequency, k is the wave number and λ is the wavelength.

#### Page No 323:

(a) must be an integral multiple of $\lambda /4$

A standing wave is produced on a string clamped at one end and free at the other. Its fundamental frequency is given by
$\nu =\left(n+\frac{1}{2}\right)\frac{v}{2L}\phantom{\rule{0ex}{0ex}}⇒v=\nu \lambda \phantom{\rule{0ex}{0ex}}⇒\nu =\left(n+\frac{1}{2}\right)\frac{\nu \lambda }{2L}\phantom{\rule{0ex}{0ex}}⇒L=\left(\frac{2n+1}{4}\right)\lambda \phantom{\rule{0ex}{0ex}}⇒L=\frac{\lambda }{4},\frac{3\lambda }{4},...$

#### Page No 323:

(b) The energy of any small part of a string remains constant in a standing wave.

A standing wave is formed when the energy of any small part of a string remains constant. If it does not, then there is transfer of energy. In that case, the wave is not stationary.

#### Page No 323:

(c) the alternate antinodes vibrate in phase
(d) all the particles between consecutive nodes vibrate in phase

All particles in a particular segment between two nodes vibrate in the same phase, but the particles in the neighbouring segments vibrate in opposite phases, as shown below. Thus, particles in alternate antinodes vibrate in the same phase.

#### Page No 324:

Given,
Pulse travelling on a string, $y=\left[\frac{{\left(a\right)}^{3}}{{\left(x-\nu t\right)}^{2}+{a}^{2}}\right]\phantom{\rule{0ex}{0ex}}$

So, at .
Similarly, at t = 1 s,

To sketch the shape of the string, we have to plot a graph between y and x at different values of t.

#### Page No 324:

Given,
Equation of the wave travelling in the positive x-direction at x = 0:

Here,
Wave speed = v
Wavelength, λ = vT
T = Time period
Therefore, the general equation of the wave can be represented by
$y=A\mathrm{sin}\left[\left(\frac{t}{T}\right)-\left(\frac{x}{\nu T}\right)\right]$

#### Page No 324:

The shape of the string at t = 0 is given by g(x) = A sin(x/a), where A and a are constants.
Dimensions of A and a are governed by the dimensional homogeneity of the equation g(x) = A sin(x/a).
Now,

#### Page No 324:

Given,
Wave velocity = $\nu$
Shape of the string at $t={t}_{0}$ =     ...(i)
For a wave travelling in the positive x-direction, the general equation is given by

Putting t = − t and comparing with equation (i), we get:

#### Page No 324:

Given,
Equation of the wave,
The general equation is $y=A\mathrm{sin}\left\{\left(\frac{2\pi x}{\lambda }\right)+\omega t\right\}$.
From the above equation, we can conclude:

(a) The wave is travelling in the negative x-direction.

(b)

Wave speed:

(c) Maximum displacement, A = 0.10 mm

#### Page No 324:

A wave travels along the positive x-direction.
Wave amplitude (A) = 0.20 cm
Wavelength (λ) = 20 cm
Wave speed (v) = 20 m/s

(a) General wave equation along the x-axis:

Wave equation:

(b) As per the question
For the wave equation ,we need to find the displacement and velocity at x = 2 cm and t = 0.

If the wave equation is written in a different fashion, then also we will get the same values for these quantities.

#### Page No 324:

The wave equation is represented by

Let:
Time period = T
Wavelength = λ

(b) Equation for the velocity of the particle:

(c) (i) Speed of the particle:
$\nu =-0.50\mathrm{cos}2\pi \left\{\frac{3}{4}-\frac{1}{2}\right\}=0$.
(ii)

(By changing the value of t, the other two can be calculated.)

#### Page No 324:

Time taken to reach from the mean position to the extreme position, $\frac{T}{4}$ = 5 ms
Time period (T) of the wave:

Wavelength (λ) =

#### Page No 324:

Given:
Wave speed,
From the graph, we can infer:
(a) Amplitude, A = 1 mm
(b) Wavelength, λ = 4 cm

(c) Wave number, $k=\frac{2\pi }{\lambda }$

(d)

#### Page No 324:

Given,
Wave speed (v) = 10 ms−1
Time period (T) = 20 ms

(a) Wavelength of the wave:

(b) Displacement of the particle at a certain instant:
$y=a\mathrm{sin}\left(\omega t-kx\right)\phantom{\rule{0ex}{0ex}}⇒1.5=a\mathrm{sin}\left(\omega t-kx\right)$
Phase difference of the particle at a distance x = 10 cm:

#### Page No 324:

Given,
Length of the steel wire = 64 cm
Weight = 5 g
Applied force = 8 N
Thus, we have:

#### Page No 324: Given,
Length of the string = 20 cm
Linear mass density of the string = 0.40 g cm−1
Applied tension = 16 N =
Velocity of the wave:

∴ Time taken to reach the other end
Time taken to see the pulse again in the original position

(b) At t = 0.01 s, there will be a trough at the right end as it is reflected.

#### Page No 324:

Given,
Linear mass density of the string = 0.5 gcm−1
Total length of the string = 30 cm
Speed of the wave pulse = 20 cms−1 The crest reflects the crest here because the wave is travelling from a denser medium to a rarer medium.

(a)
Time taken to regain shape:

(b) The wave regain its shape after covering a period distance$=2×30=60$ cm

(c) Frequency,
We know:
$n=\frac{1}{2l}\sqrt{\left(\frac{T}{m}\right)}$
Here, T is the tension in the string.
Now,

#### Page No 324:

Let:
m = Mass per unit length of the first wire
a = Area of the cross section
ρ = Density of the wire
T = Tension
Let the velocity of the first string be v1.
Thus, we have:
${\nu }_{1}=\sqrt{\left(\frac{T}{{m}_{1}}\right)}$
The mass per unit length can be given as

Let the velocity of the first string be v 2.
Thus, we have:

Given,

#### Page No 325:

Given,
Wave equation,

Velocity of the wave in the stretched string is given by

So, the tension in the string is 0.108 N.

#### Page No 325:

Given,
Amplitude of the wave = 1 cm
Frequency of the wave,
Mass per unit length, m = 0.1 kg/m
Applied tension, T = 90 N

(a) Velocity of the wave is given by
$v=\sqrt{\frac{T}{m}}$
Thus, we have:

(b) At x = 0, displacement is maximum.
Thus, the wave equation is given by
...(1)

(c) Using $\mathrm{cos}\left(-\theta \right)=\mathrm{cos}\theta$ in equation (1), we get:
$y=1\mathrm{cos}2\pi \left(\frac{x}{30}-\frac{t}{0.01}\right)$

Now,

In magnitude, v = 5.4 m/s.
Similarly,

#### Page No 325:

Given,
Length of the string, L = 40 cm
Mass of the string = 10 gm
Mass per unit length
Spring constant, k = 160 N/m

∴ Time taken by the pulse to reach the spring,

#### Page No 325:

Given,
Mass of each block =
Linear mass density of wire AB = 10 gm−1 = 0.01 kgm−1
Linear mass density of wire CD = 8 gm−1 = 0.008 kgm−1
For string CD, velocity is defined as $v=\sqrt{\left(\frac{T}{m}\right)}$.
Here, T is the tension and m is the mass per unit length.
For string CD,
$\mathrm{T}=3.2×g$
Thus, we have:

For string AB,

#### Page No 325:

Given,
Mass of the block = 2 kg
Total length of the string = 2 + 0.25 = 2.25 m
Mass per unit length of the string: Time taken by the disturbance to reach the pulley:

#### Page No 325:

Given,
Mass of the block = 4.0 kg
Linear mass density,
From the free body diagram, #### Page No 325:  Given,
Speed of the transverse pulse when the car is at rest, v1 = 60 cm s−1
Speed of the transverse pulse when the car accelerates, v2= 62 cm s−1
Let:
Mass of the heavy ball suspended from the ceiling = M
Mass per unit length = m
Now,

When car is having acceleration:
Tension, $T=\sqrt{{\left(Ma\right)}^{2}+{\left(Mg\right)}^{2}}$

From equations (i) and (ii), we get:

#### Page No 325:

Let,
V = Linear velocity of the string
m = Mass per unit length of the the string.
R = Radius of the loop
ω = Angular velocity Consider one half of the string, as shown in the figure.
The half loop experiences centrifugal force at every point (away from the centre) balanced by tension 2T.
Consider an element of angular part at angle θ.
So,
Length of the element
Centrifugal force experienced by the element
Resolving the centrifugal force into rectangular components,
Since the horizontal components cancel each other, the net force on the two symmetric elements is given as

Velocity of the transverse vibration is given as
$V\text{'}=\sqrt{\left(\frac{T}{m}\right)}\phantom{\rule{0ex}{0ex}}V\text{'}=\sqrt{\left(\frac{m{R}^{2}{\omega }^{2}}{m}\right)}=\omega R$
Linear velocity of the string, V = $\omega R$
∴ Speed of the disturbance, V' =  V

#### Page No 325:

(a) Let m be the mass per unit length of the string.
Consider an element at a distance x from the lower end.
Here,
Weight acting downwards = (mx)g
∴ Tension in the string at the upper part = mgx
The velocity of transverse vibration is given as
$v=\sqrt{\left(\frac{T}{m}\right)}=\sqrt{\left(\frac{mgx}{m}\right)}\phantom{\rule{0ex}{0ex}}⇒v=\sqrt{\left(gx\right)}$

(b) Let the time taken be dt for the small displacement dx.
Thus, we have:

$dt=\frac{dx}{v}=\frac{dx}{\sqrt{\left(gx\right)}}$

(c) Suppose after time t, the pulse meets the particle at a distance y from the lower end of the rope.

Now,

∴ Distance travelled by the particle in this time, S = $L-y$
Using the equation of motion, we get:

Thus, the particle will meet the pulse at a distance $\frac{\mathrm{L}}{3}$ from the lower end.

#### Page No 325:

Given,
Linear density of each of two long strings A and B, m =
String A is stretched by tension Ta= 4.8 N.
String B is stretched by tension Tb= 7.5 N.
Let va and vb be the speeds of the waves in strings A and B.
Now,

Distance travelled by the wave in 0.02 s in string A:
s
Relative speed between the wave in string A and the wave in string B, v'
Time taken by the wave in string B to overtake the wave in string A = Time taken by the wave in string B to cover 0.4 m

#### Page No 325:

Given,
Amplitude of the transverse wave, r = 0.5 mm
Frequency, f = 100 Hz
Tension, T = 100 N
Wave speed, v = 100 m/s
Thus, we have:

#### Page No 325:

Given,
Frequency of the wave, f = 200 Hz
Amplitude, A = 1 mm = 10−3 m
Linear mass density, m = 6 gm−3
Applied tension, T = 60 N
Now,
Let the velocity of the wave be v.
Thus, we have:

(a) Average power is given as

(b) Length of the string = 2 m
Time required to cover this distance:

#### Page No 325:

Given,
Frequency of the tuning fork, f = 440 Hz
Linear mass density, m = 0.01 kgm−1
Applied tension, T = 49 N
Amplitude of the transverse wave produce by the fork = 0.50 mm
Let the wavelength of the wave be $\lambda$.

(a) The speed of the transverse wave is given by
$\nu =\sqrt{\left(\frac{T}{m}\right)}$

(b) Maximum speed (vmax) and maximum acceleration (amax):

(c) Average rate (p) is given by

#### Page No 325:

Given,
Phase difference between the two waves travelling in the same direction, $\varphi =90°=\frac{\pi }{2}$.
Frequency f and wavelength $\lambda$ are the same. Therefore, ω will be the same.
Let the wave equations of two waves be:
${y}_{1}=r\mathrm{sin}\omega t$
${y}_{2}=r\mathrm{sin}\left(\omega t+\frac{\pi }{2}\right)$
Here, r is the amplitude.
From the principle of superposition, we get:

∴ Resultant amplitude,

#### Page No 325: Given,
Speed of the wave pulse travelling in the opposite direction, v = 50 cm s−1 = 500 mm s−1
Distances travelled by the pulses:
Using s = vt, we get:

The shapes of the string at different times are shown in the figure.

#### Page No 326:

Given:
Two waves have same frequency (f), which is 100 Hz.
Wavelength (λ) = 2.0 cm

(a) First wave will travel the distance in 0.015 s.

This will be the path difference between the two waves.
So, the corresponding phase difference will be as follows:

(b) Path difference between the two waves, x = 4 cm = 0.04 m
So, the corresponding phase difference will be as follows:

(c) The waves have same frequency, same wavelength and same amplitude.
Let the wave equation for the two waves be as follows:

∴ Resultant amplitude

#### Page No 326:

Length of a stretched string (L) = 1 m
Wave speed (v) = 60 m/s
Fundamental frequency (f0) of vibration is given as follows:
${f}_{0}=\frac{v}{2L}$

#### Page No 326:

Given:
Length of the wire (L)= 2.00 m
Fundamental frequency of the vibration (f0) = 100 Hz
Applied tension (T) = 160 N

So, the linear mass density of the wire is 1 g/m.

#### Page No 326:

Given:
Mass of the steel wire = 4.0 g
Length of the steel wire = 80 cm = 0.80 m
Tension in the wire = 50 N
Linear mass density (m)

#### Page No 326:

Given:
Length of the piano wire (L)= 90.0 cm = 0.90 m
Mass of the wire = 6.00 g = 0.006 kg
Fundamental frequency (fo) = 261.63 Hz

Hence, the tension in the piano wire is 1480 N.

#### Page No 326:

Given:
Length of the sonometer wire (L) = 1.50 m
Let the
According to the question, When the fundamental wave is produced, we have:

#### Page No 326:

Given:
Length of the wire between two pulleys (L) = 1.5 m
Mass of the wire = 12 gm Fundamental frequency is given by:

For second harmonic (when two loops are produced):

#### Page No 326:

Given:
Length of the stretched string (L) = 1.00 m
Mass of the string =40 g
String is attached to the tuning fork that vibrates at the frequency (f) = 128 Hz
Linear mass density (m)
No. of loops formed, (n) = 4

Hence, the tension in the string if it is to vibrate in four loops is 164 N.

#### Page No 326:

Given:
Wire makes a resonant frequency of 240 Hz and 320 Hz when its both ends are fixed.
Therefore, fundamental frequency (f0) of the wire must be the factor of 240 Hz and 320 Hz.
(a) Maximum value of fundamental frequency, f0 = 80 Hz

(b) Wave speed (v) = 40 m/s
And if
$\frac{\lambda }{2}=L$

#### Page No 326:

Given:
Separation between two consecutive nodes when the string vibrates in resonant mode = 2.0 cm
Let there be 'n' loops and $\lambda$ be the wavelength.
$\lambda$ =  Length of the wire is L.

In the first case:

In the second case:

∴ Length of the string,

#### Page No 326:

Given:
Frequency (f) = 660 Hz
Wave speed (v) = 220 m/s

(a) No. of loops, n = 3
∴ $L=\frac{n}{2}\lambda$

(b) Equation of resultant stationary wave can be given by:

#### Page No 326:

Given:
Length of the guitar wire (L1) = 30.0 cm = 0.30 m
Frequency, when no finger is placed on it, (f1) =196 Hz
And (f2) =220 Hz, (f3) = 247 Hz, (f4) = 262 Hz and (f5) = 294 Hz
The velocity is constant for a medium.
We have:
$f\propto \left(\frac{1}{L}\right)$

Again,

#### Page No 326:

Given:
Fundamental frequency (f0) of the steel wire = 200 Hz
Let the highest harmonic audible to the person be n.

Frequency of the highest harmonic, f' = 14000 Hz
f'=nf0    ...(1)

Thus, the highest harmonic audible to man is the 70th harmonic.

#### Page No 326:

Given:
Let the three resonant frequencies of a string be

(a) So, the highest possible fundamental frequency of the string is , because f1, f2 and f3 are the integral multiples of 30 Hz.

(b) So, these frequencies can be written as follows:
${f}_{1}=3f\phantom{\rule{0ex}{0ex}}{f}_{2}=5f\phantom{\rule{0ex}{0ex}}{f}_{3}=7f$

Hence, f1, f2, and f3 are the third harmonic, the fifth harmonic and the seventh harmonic, respectively.
(c) The frequencies in the string are f, 2f, 3f, 4f, 5f ...
∴ 3f = 2nd overtone and 3rd harmonic
5f = 4th overtone and 5th harmonic
7th= 6th overtone and 7th harmonic

(d) Length of the string (L) = 80 cm = 0.8 m
Let the speed of the wave be v.
So, the frequency of the third harmonic is given by:

#### Page No 326:

Given:
The tensions in the two wires are in the ratio of 2:1.
$⇒\frac{{T}_{1}}{{T}_{2}}=2$
Ratio of the radii is 3:1.
$⇒\frac{{r}_{1}}{{r}_{2}}=3=\frac{{D}_{1}}{{D}_{2}}$
Density in the ratios of 1:2.
$⇒\frac{{\rho }_{1}}{{\rho }_{2}}=\frac{1}{2}$
Let the length of the wire be L.

$⇒{f}_{1}=\frac{1}{L{D}_{1}}\sqrt{\frac{{T}_{1}}{\pi {\rho }_{1}}}\phantom{\rule{0ex}{0ex}}⇒{f}_{2}=\frac{1}{L{D}_{2}}\sqrt{\frac{{T}_{2}}{\pi {\rho }_{2}}}$

#### Page No 326:

Given:
Length of the rod (L) = 40 cm = 0.40 m
Mass of the rod (m) = 1.2 kg
Let the mass of 4.8 kg be placed at x distance from the left.
As per the question, frequency on the left side = f0
Frequency on the right side = 2f0
Let tension be T1 and T2 on the left and the right side, respectively.

From the free body diagram: Now, taking moment about point A: Therefore, the mass should be placed at a distance of 5 cm from the left end.

#### Page No 326:

Given:
Length of the aluminium wire (La)= 60 cm = 0.60 m
Length of the steel wire (Ls)= 80 cm = 0.80 m
Tension produced (T) = 40 N
Area of cross-section of the aluminium wire (Aa)  = 1.0 mm2
Area of cross-section of the steel wire (As) = 3.0 mm2
Density of aluminium (ρa) = 2⋅6 g cm−3
Density of steel  (ρs) = 7⋅8 g cm−3 A node is always placed at the joint. Since aluminium and steel rod has same mass per unit length, the velocity of wave (v) in both of them is same.
Let v be the velocity of wave.

For minimum frequency, there would be maximum wavelength.
For maximum wavelength, minimum number of loops are to be produced.
∴ Maximum distance of a loop = 20 cm

#### Page No 326:

Given:
Length of the string = L
Velocity of wave is given as:
$V=\sqrt{\frac{T}{m}}$
In fundamental mode, frequency = ν
$⇒\nu =\frac{1}{2L}\sqrt{\frac{T}{m}}$
(a)

(b) Equation of the stationary wave:

#### Page No 327:

Given:
Length of the string (L) = 2.0 m
Wave speed on the string in its first overtone (v) = 200 m/s
Amplitude (A) = 0.5 cm

(a) Wavelength and frequency of the string when it is vibrating in its 1st overtone (n = 2):
$L=\frac{n\lambda }{2}$

(b) The stationary wave equation is given by:

#### Page No 327:

Given:
The stationary wave equation of a string vibrating in its third harmonic is given by
y = (0.4 cm) sin [(0.314 cm−1) x]cos [(.600 πs−1) t]
By comparing with standard equation,

(a) From the above equation, we can infer the following:

Wavelength, $\lambda =\frac{2\pi }{0.314}=\frac{\left(2×3.14\right)}{0.314}$

(b) Therefore, the nodes are located at 0cm, 10 cm, 20 cm, 30 cm. (c) Length of the string, l = $n\frac{\lambda }{2}$

(d)

$\lambda$ and $\nu$ are the wavelength and velocity of the waves that interfere to give this vibration.

#### Page No 327:

Given:
Equation of the standing wave:

We know:
$L=\frac{n\lambda }{2}$
For the smallest length, putting n = 1:

Therefore, the required length of the string is 10 cm.

#### Page No 327:

Given:
Length of the wire (L) = 40 cm = 0.40 m
Mass of the wire = 3.2 g = 0.003 kg
Distance between the two fixed supports of the wire = 40.05 cm
Fundamental mode frequency = 220 Hz
Therefore, linear mass density of the wire (m) is given by:

Hence, the required Young's modulus of the wire is .

#### Page No 327:

Density of the block = ρ
Volume of block = V
∴ Weight of the block is, W = ρVg
∴ Tension in the string, T = W

The tuning fork resonates with different frequencies in the two cases.
Let the tenth harmonic be f10. Here, m is the mass per unit length of the string and L is the length of the string.

When the block is immersed in water (density = ρ ), let the eleventh harmonic be f11. The frequency (f) of the tuning fork is same.

Therefore, the required density is .

#### Page No 327:

Given:
Length of the long rope (L) = 2.00 m
Mass of the rope = 80 g = 0.08 kg
Tension (T) = 256 N
Linear mass density, m

For fundamental frequency:

(a) The frequency overtones are given below:

(b)

Hence, the wavelengths are 8 m, 2.67 m and 1.6 m, respectively.