NCERT Exemplar Class 11 Physics Solutions Chapter 15 Waves

Question:15.2

Sound waves of wavelength $\lambda$ travelling in a medium with a speed of v m/s enter into another medium where its speed is 2v m/s. The wavelength of sound waves in the second medium is
$(a) \lambda$
$(b)\frac{ \lambda}{2}$
$(c) 2\lambda$
$(d) 4\lambda$

Answer:

The answer is the option (c)
$v= u \lambda , u= \frac{v}{\lambda}$
When a wave passes from one medium to another, its frequency does not change but its velocity and wavelength changes.
$\frac{v}{\lambda} = \frac{2v}{\lambda _{2}}$
Thus, $\lambda _{2}=2 \lambda$

Question:15.3

Speed of sound wave in the air
(a) is independent of temperature.
(b) increases with pressure.
(c) increases with increase in humidity.
(d) decreases with increase in humidity.

Answer:

The answer is the option (c) Sound is a longitudinal wave with speed
$v = \sqrt{\frac{\lambda P}{\rho}}$ .
The density of water vapours is less than the air so, on increasing humidity, the density of medium decreases, in turn, increasing the speed of sound in air by
$v \propto \frac{1}{\rho }$ .
Hence, c is correct.

Question:15.4

Change in temperature of the medium changes
(a) frequency of sound waves.
(b) the amplitude of sound waves.
(c) the wavelength of sound waves.
(d) the loudness of sound waves.

Answer:

The answer is the option (c)
Thus, on increasing temperature, the speed also increases, as frequency remains constant during the propagation of the wave. Since $v=u\lambda$, the wavelength also increases with velocity.

Question:15.5

With the propagation of longitudinal waves through a medium, the quantity transmitted is
(a) matter.
(b) energy.
(c) energy and matter.
(d) energy, matter, and momentum.

Answer:

The answer is the option (b) During propagation of a wave in any medium, only energy is transmitted from one point to another and not matter.

Question:15.6

Which of the following statements are true for wave motion?
(a) Mechanical transverse waves can propagate through all mediums.
(b) Longitudinal waves can propagate through solids only.
(c) Mechanical transverse waves can propagate through solids only.
(d) Longitudinal waves can propagate through a vacuum.

Answer:

The answer is the option (c) A mechanical transverse wave can propagate through solid and on the surface of the liquid as well. A transverse wave cannot reproduce in gases whereas Longitudinal wave can. Also, longitudinal waves are not electromagnetic waves and can never propagate in a vacuum.

Question:15.7

A sound wave is passing through the air column in the form of compression and rarefaction. In consecutive compressions and rarefactions,
a) density remains constant
b) Boyle’s law is obeyed
c) bulk modulus of air oscillates
d) there is no transfer of heat

Answer:

The answer is the option (d) In consecutive compressions and rarefactions, the density of medium particles is maximum and minimum, respectively. Due to the rapid change in density temperature rises while the bulk modulus remains constant. The time of compressions and rarefactions is very small, so heat does not transfer.

Question:15.8

Equation of a plane progressive wave is given by $y = 0.6 \sin 2\pi\left ( t-\frac{x}{2} \right )$. On reflection from a denser medium, its amplitude becomes $\frac{2}{3}$ of the amplitude of the incident wave
a) $y = 0.6 \sin 2\pi\left ( t+\frac{x}{2} \right )$
b) $y = -0.4 \sin 2\pi\left ( t+\frac{x}{2} \right )$
c) $y = 0.4 \sin 2\pi\left ( t+\frac{x}{2} \right )$
d) $y = -0.4 \sin 2\pi\left ( t-\frac{x}{2} \right )$

Answer:

The answer is the option b)
$y = -0.4 \sin 2\pi\left ( t+\frac{x}{2} \right )$
After reflection of wave changes by phase $180^{\circ}$
$y_{r}=\left (\frac{2}{3}\times 0.6 \right )\sin2\pi \left [\pi +t+\frac{x}{2} \right ]$
$y_{r} = -0.4 \sin 2\pi\left ( t+\frac{x}{2} \right )$

Question:15.9

A string of mass 2.5 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in
(a) one second
(b) 0.5 second
(c) 2 seconds
(d) data given is insufficient

Answer:

The answer is the option (b) M= mass of string= 2.5Kg, l=20m
μ=mass per unit length= $\frac{M}{l}=\frac{2.5}{20} = 0.125 kg/m$
$v=\sqrt{ \frac{T}{\mu }}=\sqrt{\frac{200}{0.125}}=\sqrt{1600}=\frac{40m}{s}$
Time =
$\frac{distance }{speed}= \frac{20m}{\frac{40m}{s}}=\frac{1}{2}sec=0.5 sec$

Question:15.10

A train whistling at constant frequency is moving towards a station at a constant speed V. The train goes past a stationary observer on the station. The frequency n′ of the sound as heard by the observer is plotted as a function of time t (Figure). Identify the expected curve.

Answer:

The answer is option (c) When the observer is at rest and the source of sound is moving towards the observer, then the observed frequency
$n'=\left (\frac{v}{v-v_{s}} \right )n_{0}$ where n₀ is the original frequency of the source of sound
v=speed of sound in the medium
$n'>n_{0}$
v_s=speed of source
When the source is moving away from the observer
$n" = \frac{vn_{0}}{v+v_{s}}$
$n^{''}< n_{0}$
Thus, n’ > n” and graph (c) is correct.

Question:15.12

The displacement of a string is given by
$y (x,t) = 0.06 \sin \left (\frac{2 \pi x}{3} \right ) \cos (120\pi t)$
where x and y are in m and t in s. The length of the string is 1.5m and its mass is $3.0 \times 10^{-2}$ kg
(a) It represents a progressive wave of frequency 60Hz.
(b) It represents a stationary wave of frequency 60Hz.
(c) It is the result of superposition of two waves of wavelength 3 m, frequency 60Hz each travelling with a speed of 180 m/s in opposite direction.
(d) Amplitude of this wave is constant.

Answer:

The answer is the option (b) and (c)
Equation of stationary wave is $y(x,t)=a \sin\left (kx \right )\cos\left (\omega t \right )$
Since the waves are stationary, the amplitude varies from 0 to a=0.06m from nodes to antinodes.
Comparing this equation with the given equation $y (x,t) = 0.06 \sin \left (\frac{2 \pi x}{3} \right ) \cos (120\pi t)$
We get$\omega =120\pi =2\pi \nu ; \nu=60Hz$
From equation $\frac{2\pi }{3}=k=\frac{2\pi }{\lambda}$
$\lambda =3m, \nu =60Hz$
Speed = $60\times 3=180m/s$

Question:15.13

Speed of sound waves in a fluid depends upon
(a) directly on density of the medium.
(b) square of Bulk modulus of the medium.
(c) inversely on the square root of density.
(d) directly on the square root of bulk modulus of the medium.

Answer:

The answer is the option (c) and (d) Speed of sound wave in the fluid of bulk modulus K and density $\rho$ is given by
$v=\sqrt{\frac{K}{\rho}}$
$v \propto \sqrt{K}$ (if $\rho$ is constant)
$v\propto\sqrt{\frac{1}{\rho}}$ (if K is constant)

Question:15.14

During propagation of a plane progressive mechanical wave
(a) all the particles are vibrating in the same phase.
(b) amplitude of all the particles is equal.
(c) particles of the medium executes S.H.M.
(d) wave velocity depends upon the nature of the medium.

Answer:

The answer is the option (b) (c) and (d) During propagation of a mechanical wave, each particle is displaced from 0 to its amplitude (which is equal for each particle). Each particle between any 2 successive crests and roughs are in a different phase.
In case of a progressive wave, medium particles oscillate about their mean position with a restoring force$(F \propto (-y))$acting on them like in a simple harmonic motion.
Also, $v=\sqrt{\frac{K}{\rho}}$Where K and $\rho$depend on the nature of the medium.

Question:15.15

The transverse displacement of a string is given by $y(x,t) = 0.06 \sin \left (\frac{2\pi x}{3} \right )\cos (120\pi t)$. All the points on the string between two consecutive nodes vibrate with
a) same frequency
b) same phase
c) same energy
d) different amplitude

Answer:

The answer is the option (a), (b), and (d).
The given equation $y(x,t) = 0.06 \sin \left (\frac{2\pi x}{3} \right )\cos (120\pi t)$ is as per the standard equation of stationary wave, i.e. $y(x,t)=a \sin kx\cos\omega t$

The frequency of particles within the wave is the same. All the particles between any two consecutive nodes vibrate either the upside or downside having the same phase $120\pi t$ at a time.
Also, $E \propto A^{2}$ where the amplitude of different particles is different between nodes.

Question:15.16

A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m/s. Given that the speed of sound in still air is 340 m/s
a) the frequency of sound as heard by an observer standing on the platform is 400 Hz
b) the speed of sound for the observer standing on the platform is 350 m/s
c) the frequency of sound as heard by the observer standing on the platform will increase
d) the frequency of sound as heard by the observer standing on the platform will decrease

Answer:

The answer is the option (a) and (b) Frequency of source of sound $v_{0}=400 Hz$
The velocity of wind from source
$v_{w}=\frac{10m}{s}$
Speed of sound in still air
$v=\frac{340m}{s}$
Now, the speed of sound with respect to the listener
$v+v_{w}=340+10=\frac{350m}{s}$
As the distance between listener and source does not change, so the frequency of sound does not change as heard by the listener.

Question:15.17

Which of the following statements are true for a stationary wave?
a) every particle has a fixed amplitude which is different from the amplitude of its nearest particle
b) all the particles cross their mean position at the same time
c) all the particles are oscillating with the same amplitude
d) there is no net transfer of energy across any plane
e) there are some particles which are always at rest

Answer:

The answer is the option (a) (b) (d) and (e)
In a stationary wave, the particles between two nodes vibrate with different amplitude, which increases from node to antinode (0 to maximum) and decreases from anti-node to node. The amplitude of particles varies with $\lambda$. The particles at a node are at rest, and hence there is no net transfer of energy. The particle between two nodes is in the same phase. The motion of particles between two nodes will be either upward or downward crossing the mean position at the same time. Hence, a,b,d and e are correct.

Question:15.19

An organ pipe of length L open at both ends is found to vibrate in its first harmonic when sounded with a tuning fork of 480 Hz. What should be the length of a pipe closed at one end, so that it also vibrates in its first harmonic with the same tuning fork?

Answer:

As the medium, number and frequency of harmonic in open and closed pipes are the same, so the number of nodes and wavelength in both cases will also be the same.
In both end open pipe
$L_{1}=2\times \frac{\lambda _{1}}{4} or\lambda _{1}=2L_{1}$
$v_{1}=\frac{c}{\lambda _{1}}=\frac{c}{2L_{1}}$
In one open end pipe
$L_{2}=1\times \frac{\lambda _{2}}{4} or\lambda _{2}=4L_{2}$
$v_{2}=\frac{c}{4L_{2}}$
$v_{1}=v_{2} and c_{1}=c_{2}=c$

$\frac{c}{2L_{1}}=\frac{c}{4L_{2}}$
Or $\\4L_{2}=2L_{1} ;\\L_{2}=L_{1}/2$

Question:15.2

A tuning fork A, marked 512 Hz, produces 5 beats per second, where sounded with another unmarked tuning fork B. If B is loaded with wax the number of beats is again 5 per second. What is the frequency of the tuning fork B when not loaded?

Answer:

$v_{A}=512$
$v_{o}=v_{A}-v_{B}$
On loading frequencies of B decreases
$v_{1}=5$
$v_{B}=v_{A}\pm 5$
$v_{B}=512\pm 5=507 or 517$
Therefore, on loading the frequency of B decreases to 507
Frequency of tuning fork, when unloaded, is 517 Hz.
(When frequency decreases by 10 Hz the number of beats will be same as 512-507=5)

Question:15.21

The displacement of an elastic wave is given by the function $y = 3 \sin \omega t + 4 \cos \omega t$ where y is in cm and t is in second. Calculate the resultant amplitude.

Answer:

$y=3 \sin\omega t+4\cos \omega t$
Let $3=a\cos\phi and 4=a \sin\phi ;$
$\tan\phi =\frac{4}{3} or \phi = tan^{-1}\frac{4}{3}$
$a^{2}(\cos\phi )^{2}+a^{2}(\sin\phi) ^{2}=3^{2}+4^{2}$
$a^{2}(\cos^{2}\phi +\sin^{2}\phi )=9+16$
$a^{2} = 25$
a=5
Now, $y=a\cos\phi \sin \omega t+a\sin\phi \cos\omega t$
$y=a \sin (\omega t+\phi)$
$y=5\sin (\omega t+\phi)$ where
$\phi = tan^{-1}\frac{4}{3}$

Question:15.22

A sitar wire is replaced by another wire of same length and material but of three times earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?

Answer:

Frequency of wire stretched at both ends $v=\frac{n}{2L}\sqrt{\frac{T}{m}}$
As the number of harmonics, length L and tension T is kept the same in both cases $v\propto \frac{1}{\sqrt{m}}$
$v_{1}/v_{2}=\sqrt{\frac{m_{2}}{m_{1} }}$
mass per unit length=mass of wire/length=$\frac{\pi r^{2}l\rho}{l} = (\pi r^{2})\rho$
As the material of wire is same
$\frac{m_{2}}{m_{1}}=\frac{(\pi r^{2}_{2})\rho}{(\pi r_{1}^{2})\rho} =\frac{9}{1}$
$\frac{v_{1}}{v_{2}}=\sqrt{\frac{9}{1}} =3$

Question:15.23

At what temperatures will the speed of sound in air be 3 times its value at $0^{\circ}C$?

Answer:

We know that
$v \propto T$
Let v_T be speed of sound in air at temp T and v₀ at $0^{\circ}C$
$\frac{ v_{T}}{v_{0}}=\sqrt{\frac{T}{T_{0}}}$ and $v_{T}=3v_{0}$

$\frac{3v_{0}}{v_{0}}=\sqrt{\frac{T}{273+0 }}$

Or $T = 2457K = 2184^{\circ}C$

Question:15.24

When two waves of almost equal frequencies n₁ and n₂ reach at a point simultaneously, what is the time interval between successive maxima?

Answer:

Since the frequencies of vibrations are almost equal $n_{1}=n_{2}$
Assuming that $n_{1}>n_{2}$ number of beats per second=$n =n_{2}-n_{1}$
So time period of maxima or beats
$\frac{1}{n}=\frac{1}{n_{2}-n_{1}}$

Question:15.26

A pipe of 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz?

Answer:

$l=20 cm=0.2m, v=1237.5 Hz,$ $V=\frac{330m}{s}$
$l=\frac{\lambda}{4}$
For fundamental frequency v
$\nu_{1}=\frac{v}{\lambda }=\frac{v}{4l}=\frac{330}{4\times 20\times 10^{-2}}=412.5 Hz$

$\\v=1237.5 Hz \\\\\frac{ v}{v_{1}}=\frac{1237.5}{412.5}=31$
The 3^rd harmonic is excited by 1237.5 Hz

Question:15.27

A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz still air. The train beings to move with a speed of 10 m/s towards the platform. What is the frequency of the sound for an observer standing on the platform?

Answer:

$v_{0}=400Hz$ and

$V_{s}=\frac{10m}{s}$given
Velocity of sound in air
$V_{a}=\frac{330m}{s}$
Apparent frequency by the observer standing on the platform
$v'=\frac{V_{a}}{V_{a}-V_{s} }v_{0}$
$v'=330\times \frac{400}{330-10}=412.5Hz$

Question:15.28

The wave pattern on a stretched string is shown in the figure. Interpret what kind of wave this is and find its wavelength.

Answer:

The displacement of medium particles at a distance 10,20,30,40 and 50cm are always at rest. (property of nodes in stationary waves)
At$t=\frac{T}{4} and \frac{3T}{4}$ all particles are at rest which happens in a stationary wave when the particle crosses its mean position.
So, the graph of wave shows a stationary wave. Nodes are at x=10, 20,30, 40 cm and distance between successive nodes is
$\frac{\lambda}{2}=(30-20); \lambda =20cm$

Question:15.29

The pattern of standing waves formed on a stretched string at two instants of time are shown in the figure. The velocity of two waves superimposing to form stationary waves is 360 m/s and their frequencies are 256 Hz.

a) calculate the time at which the second curve is plotted
b) mark nodes and antinodes on the curve
c) calculate the distance between A’ and C’

Answer:

Frequency of wave $\nu=256Hz$
$T=\frac{1}{\nu}=\frac{1}{256}second=0.00390=3.9\times 10^{-3}seconds$

1. In stationary wave, a particle passes through its mean position after every $\frac{T}{4}$ time

$T=\frac{T}{4} =3.9\times \frac{10^{-3}}{4}=0.975\times 10^{-3} sec=9.75\times 10^{-4}sec$

2. Point does not vibrate at A, B, C, D and E. Points A’ and C’ are at maximum displacement

Between A’ and C’ = $\lambda =\frac{V}{v}=\frac{360}{256}=1.41m$

Question:15.30

A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water. The water level in the tube is gradually lowered. When the water level is 17 cm below the open end, the maximum intensity of sound is heard. If the room temperature is $20^{\circ}C$, calculate
a) speed of sound in air at room temperature
b) speed of sound in air at $0^{\circ}C$
c) if the water in the tube is replaced with mercury, will there be any difference in your observations?

Answer:

(a) Pipe partially filled with water behaves like an open organ pipe. The first harmonic is heard at L=17 cm
$L=\frac{\lambda }{4} ;\lambda =4L=0.68m$
$V=\frac{v}{\lambda}=512\times \frac{0.68m}{s}=\frac{348.16m}{s}$

(b) $\frac{V_{0}}{V_{T}}=\sqrt{\frac{T_{0}}{T} }$
$\frac{V_{0}}{348.16}=\sqrt{\frac{273}{293 }}$
$V_{0}=348.16\times 0.96526=\frac{336m}{s}$

(c) Water and mercury in the tube reflect sound into the air column to form stationary waves and the reflection is more in mercury compared to water because of its higher density. The intensity of sound heard in mercury will be larger, but the reading remains the same as the medium in tube and tuning fork does not change.

Question:15.31

Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops, and 4 loops, the frequencies are in the ratio 1:2:3:4.

Answer:

Let the number of loops in the string be n
Length of each loop is $\frac{\lambda }{2}$
$L=n\frac{\lambda}{2} or \lambda =\frac{2L}{n}$
$V=\frac{v}{\lambda} and \lambda =\frac{V}{v}=\frac{2L}{n}$
$v=\frac{Vn}{2L }$Where V in a stretched string is $\sqrt{\frac{T}{m}}$

$v=\frac{n}{2L}\sqrt{\frac{T}{m}}$
For n=1, $v_{1}=\frac{1}{2L}\sqrt{\frac{T}{m}}=v_{0}$
For n=2, $v_{2}=\frac{2}{2L}\sqrt{\frac{T}{m}}=2v_{0}$
Similarly for n=3 and so on
$v_{1}:v_{2}:v_{3}:v_{4}=n_{1}:n_{2}:n_{3}:n_{4}=1:2:3:4$

Question:15.33

If c is the r.m.s. speed of molecules in a gas and v is the speed of sound waves in the gas, show that c/v is constant and independent of temperature for all diatomic gases.

Answer:

We know that $c= \sqrt{\frac{3P}{\rho}}$ for molecules
$c= \sqrt{\frac{3RT}{M}}$
$\frac{p}{\rho}=\frac{PT}{M}$ where M is the molar mass of gas
$\frac{P}{\rho}=\frac{\frac{RT}{V}}{\frac{M}{V}}$
$v=\frac{\gamma P}{\rho}=\frac{\gamma RT}{M}$
PV=nRT When n=1
$P=\frac{RT}{V}$$cv=\sqrt{\frac{\frac{3RT}{M}}{\frac{\gamma RT}{M}}} ={\sqrt{}\frac{3}{\gamma }}$
$c/v=\sqrt{\frac{\frac{3RT}{M}}{\frac{\gamma RT}{M}}} ={\sqrt{}\frac{3}{\gamma }}$
$\gamma =\frac{C_{p}}{C_{v}}=\frac{7}{5}$adiabatic constant for diatomic gas
$\frac{c}{v}= \sqrt{\frac{15}{7}}$=constant

Question:15.34

Given below are some functions of x and t to represent the displacement of an elastic wave.
$a) y = 5 \cos (4x) \sin (20t)$
$b) y = 4 \sin\left ( 5x-\frac{t}{2} \right )+ 3 \cos \left ( 5x-\frac{t}{2} \right )$
$c) y = 10 \cos [(252-250)\pi t] \cos [(252+250) \pi t]$
$d) y = 100 \cos (100 \pi t + 0.5x)$
State which of these represent
a) a travelling wave along -x direction
b) a stationary wave
c) Beats
d) a travelling wave along +x direction
Give reasons for your answers

Answer:

(a) A wave travelling along the (-x) direction has +kx as in
$d) y = 100 \cos (100 \pi t + 0.5x)$
(b ) A stationary wave equation is
$a) y = 5 \cos (4x) \sin (20t)$

1. Beats involve $\left (v_{1}+v_{2} \right )$ and $\left (v_{1}-v_{2} \right )$ So they can be represented by

$y=\left [\cos\left (252-250 \right )\pi t \right ]$ i.e. option c

2. $y = 4 \sin\left ( 5x-\frac{t}{2} \right )+ 3 \cos \left ( 5x-\frac{t}{2} \right )$

Let $4=a\cos\phi and 3=a\sin\phi$ ;
$\tan\phi =\frac{3}{4} or \phi = tan^{-1} \frac{3}{4}$
$a^{2}\cos \phi ^{2}+a^{2}\sin \phi ^{2}=3^{2}+4^{2}$
$a^{2}(\cos^{2}\phi +sin^{2}\phi )=9+16$
$a^{2}=25$
a=5
Now, $y=a\cos \phi \sin\left (5x-\frac{t}{2} \right )+a\sin \phi \cos\left (5x-\frac{t}{2} \right )$
$y = a \sin \left (5x -\frac{t}{2} + \phi \right )$
$y = 5 \sin \left (5x -\frac{t}{2} + \phi \right )$
This represents the progressive wave in +x direction as the sign of kx or (5x) and$\omega t\left ( \frac{1}{2t} \right )$ are opposite so it travels in+x direction.

Question:15.35

In the given progressive waves $y = 5 \sin (100 \pi t -0.4 \pi x)$ where y and x are in m, t is in s. What is the
a) amplitude
b) wavelength
c) frequency
d) wave velocity
e) particle velocity amplitude

Answer:

Given the progressive wave equation
$y=a\sin \omega t-kx+\phi$
$y = 5 \sin (100 \pi t -0.4 \pi x)$
(a) Amplitude a=5m
$( b )k=\frac{2\pi }{\lambda}=0.4\pi$
Wavelength
$\lambda =\frac{2\pi}{k} =\frac{2\pi }{0.4\pi }=5m$
$(c) \omega =2\pi v;v=\frac{\omega}{2\pi} , where \omega =100\pi$
$v=\frac{100\pi}{2\pi} =50Hz$
Wave velocity
$V=v\lambda =50\times 5=\frac{250m}{s}$
(e) Particle velocity in the direction of amplitude at a distance x from the source
$y=5\sin 100\pi t-0.4\pi x$
$\frac{dy}{dt}=5\times 100\pi \cos\left (100\pi t-0.4\pi x \right )$
The maximum velocity of the particle is at its mean position
$\cos100 \pi t-0.4\pi x=1$
$100\pi t-0.4\pi x=0$
$\frac{dy}{dt}=5\times 100\pi =500 \pi m/s$

Question:15.36

For the harmonic travelling wave $y = 2 \cos 2 \pi (10t-0.0080x+3.5)$ where x and y are in cm and t is second. What is the phase difference between the oscillatory motion at two points separated by a distance of
a) 4 m
b) 0.5 m
c) $\frac{\lambda }{2}$
d) $\frac{3\lambda }{4}$
e) What is the phase difference between the oscillation of a particle located at x = 100 cm at t = Ts and t = 5s?

Answer:

$y = 2 \cos 2 \pi (10t-0.0080x+3.5)$
$y=2\ cos\left (2\pi 10t-0.016\pi x+7\pi \right )$
Wave is propagated in +x direction because $\omega$t and kx are in opposite sign in
standard equation $y=a\cos\omega t-kx+\phi$
$a=2, \omega =20\pi , k=0.016\pi and \phi =7\pi$
apath difference=4m=400cm
phase difference $\Delta \phi =\frac{2\pi }{\lambda } \times p=\frac{2\pi}{ \lambda} \times 400=k\times 400=0.016\pi \times 400 = 0.64 \pi rad$
(b) path difference=0.5 m
$\Delta \phi =\frac{2\pi}{\lambda} \times p=k\times p=0.016\pi \times 50=0.8 \pi rad$
(c ) path difference=$\lambda _{2}$
$\Delta \phi =\frac{2\pi }{\lambda} \times p=\frac{2\pi}{ \lambda} \times \frac{\lambda}{2} =\pi radian$
$d) \Delta \phi =\frac{2\pi}{\lambda} \times p=\frac{2\pi }{\lambda} \times \frac{3\lambda}{4} =\frac{3}{2}\pi radian$
(e) $T = \frac{2\pi}{\omega} =\frac{2\pi}{20\pi } =\frac{1}{10}sec$
At x=100 , t=T
$\phi _{1}=20\pi T-0.016\pi( 100)+7\pi =20\pi \times\frac{ 1}{10}-1.67\pi +7\pi =7.4\pi$
At t=5s
$\phi _{2}=20\pi (5)-0.016\pi 100+7\pi =100\pi -1.67\pi +7\pi =105.4\pi$
$\phi _{2}-\phi _{1}=98\pi radian$