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CBSE Questions for Class 11 Engineering Physics Waves Quiz 16 - MCQExams.com
CBSE
Class 11 Engineering Physics
Waves
Quiz 16
A wave of frequency 400 Hz has a velocity of 320 $$ms^{-1}$$. The distance between the particles differing in phase by $$90^o$$ is
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20 cm
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40 cm
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60 cm
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30 cm
Which of the following equations does not represent a progressive wave ?
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$$y=Asin[\omega (t-\frac { x }{ v } )]$$
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$$y=Asin[ \frac { 2pi }{ \lambda }(vt-x)] $$
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$$y=Asin[2\pi (\frac {t}{T}-\frac { x }{ \lambda } )]$$
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$$y=Asin[2\pi (\frac {t}{T}-\frac { x }{ v } )]$$
The spherical waves are emitted in a medium by a source of 5 watt. Assuming that medium does not absorbed energy,the intensity of wave at a distance of 5m from source will be
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$$ 1 Watt / m^2 $$
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$$ 1 / (20 \pi ) Watt / m^2 $$
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$$ 1 / (4 / \pi) watt / m^2 $$
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$$ 1/5 watt / m^2 $$
Sinusoidal wave having frequency $$500Hz$$ and speed $$800m/s$$.If two point in the medium are separated by distance
z
having phase difference $$\frac{\pi }{6},$$ then
x
is
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$$\frac{2}{{15}}m$$
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$$\frac{1}{{15}}m$$
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$$\frac{7}{{15}}m$$
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$$\frac{8}{{15}}m$$
Consider a wave with frequency $$300\mathrm { Hz }$$ and speed $$350\mathrm { ms } ^ { - 1 } .$$ How far apart are two points $$60 ^ { \circ }$$ out of
phase?
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$$\dfrac { 1 } { 7 } m$$
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$$\dfrac { 36 } { 7 } \mathrm { m }$$
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$$\dfrac { 7 } { 36 } \mathrm m$$
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$$7 \mathrm { m }$$
The energy in the superposition of waves:
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Is lost
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Increase
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remain same, only redistribution occurs
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None of the above
Two small boats are 10 m apart on a lake. Each pops up and down with a period of 4.0 seconds due to wave motion on the surface of water. What one boat is at its highest point, the other boat is at lowest point. Both boats are always within a single cycle of the waves. The speed of the waves is :
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2.5 m/s
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5.0 m/s
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14 m/s
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40 m/s
A wave equation which gives the displacement along the Y direction is given by $$y = 10^{-4} sin (60t + 2x)$$ where x and y are in metres and t is time in seconds. This represents a wave
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traveling with a velocity of 30 m/s in the negative x direction
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of wavelength $$\pi$$ metre
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of frequency $$30/\pi$$ hertz
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of amplitude $$10^{-4}$$ metre
Source $$S$$ emits microwaves with a constant amplitude. The microwave hit a metal screen $$P$$ and are reflected. A stationary wave is formed between $$S$$ and $$P$$. The wavelength of the microwaves is much smaller than the distance between $$S$$ and$$P$$
A detector $$Q$$ is moved at a slow, constant speed from $$S$$ to $$P$$
What happens to the amplitude of the signal detected by $$Q$$>
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decreases steadily
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increases and decreases steadily
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increases steadily
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remains constant
Explanation
Since the stationary wave is formed ,so there will be Nodes and Antinodes.
We are told that the wavelength of microwave is much smaller than the distance between the S and P.
This means that the stationary wave consists of several wavelength and thus many Nodes (N) and Antinodes (A) are present.
That is like
$$N\ A\ N\ A\ N\ A\ N\ .......$$
The amplitude of the signal detected by Q would increases (at Antinodes) and decreases (at Nodes).
The displacement $$y$$ of a particle, if given by $$y=4\cos^2\left(\dfrac{t}{2}\right)\sin(1000t)$$. This expression may be considered to be a result of the superposition of how many simple harmonic motions?
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$$4$$
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$$3$$
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$$2$$
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$$5$$
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$$6$$
Explanation
$$\because$$ Equation of displacement of particle is
$$y=4\cos^2(\dfrac{t}{2})\sin (1000t)$$...........(i)
Or $$y=4[\left(\dfrac{1+\cos t}{2}\right)\sin1000t]$$
$$(\because \cos^2x=\dfrac{1+\cos 2x}{2})$$
Or $$y=3\sin 1000t+2\cos t\sin 1000 t$$
Or $$y=2\sin 1000t+\sin 1001t+\sin t 999t$$
$$[2\sin A\cos B=\sin (A+B)+\sin (A-B)]$$
Hence, equation $$(i)$$ is the result of the superposition of three simple harmonic motion.
So, the correct option will be $$(B)$$
A traveling wave is represented by the equation $$ y = \frac{1}{10} sin(60 t + 2x) $$, where x and y in meters and t is in second . this represents a wave
(1) of frequency $$ \frac {30}{\pi} Hz $$
(2) of wavelength $$ \pi m $$
(3)of amplitude 10 cm
(4) moving in the positive x direction
pick out the correct statements from the above.
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1, 2, 4
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1, 3, 4
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1, 2, 3
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all
String $$I$$ and $$II$$ identical lengths and linear mass densities but string $$I$$ is under greater tension than string $$II$$. The accompanying figure shows four different situation, $$A$$ to $$D$$ in which standing wave patterns exists on the two strings. In which situation is it possible that strings $$I$$ and $$II$$ are oscillating at the same resonant frequency?
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Explanation
As we know,
$$v\propto \sqrt T$$
and since $$T_1>T_2$$
then $$v_1>v_2$$
and for both string to have same frequency
$$\lambda _1>\lambda _2$$
this is only shown in option C
so, option (C) is correct.
Two identical wave $$A$$ and $$B$$ are produced from the origin at different instants $$t_A$$ and $$t_S$$ along the positive $$x-$$ axis, as shown in the figure. If the speed of the wave is $$5\ m/s$$ then :
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the wavelength of the waves is $$1m$$
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the amplitude of the wave is $$10\ mm$$
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the wave $$A$$ leads $$B$$ by $$0.0167\ s$$
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the wave $$B$$ leads $$A$$ by $$1.67\ s$$
Explanation
The wavelength and amplitude is clear from the diagram is 1m and 10mm.
$$v=f\lambda$$ so both have same frequency as well.
A harmonic wave has been set up on a very long string which travels along the length of the string. The wave has frequency of $$50 \,Hz,$$ amplitude $$1 \,cm$$ and wavelength $$0.5 \,m.$$ For the above-described wave.
$$Statement \ I :$$ Time taken by the wave to travel a distance of $$8 \,m$$ along the length of the string is $$0.32 \,s.$$
$$Statement \ II :$$ Time taken by a point on the string to travel a distance of $$8 \,m,$$ once the wave has reached at that point and sets it into motion is $$0.32 \,s.$$
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Both the statement are correct
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Statement I is correct but Statement II is incorrect
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Statement I is incorrect but Statement II is correct
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Both the statement are incorrect.
Explanation
The wave is travelling along the length of a string, while particles constituting the string are oscillation in a direction perpendicular to the length to string. In one time period (cycle) , the wave moves forward by one wavelength while the particle on string travels a distance of 4 times the amplitude.
Here, $$T = 1/f = 0.02 \,s$$
Wave speed, $$v = f\lambda = 25 \,m/s$$
Time taken by wave to travel a distance of 8 m.
$$t_1 = 8/25 \,s = 0.32 \,s.$$
Time taken by particle on string to travel a distance of 8 m,
$$t_2 = \dfrac {8 \times T}{4 \,times \,amplitude} = \dfrac {8}{4 \times 0.01} \times 0.02 = 4 \,s$$
The initial phase angle observed by a rider in the elevator,taking downward direction to be positive and positive extreme position to have $$\pi/2$$ phase constant, is equal to:
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zero
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$$\pi/2\ rad$$
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$$\pi/ rad$$
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$$3\pi/2\ rad$$
Explanation
As elevator was rising up with $$a=10\,m/s^2$$ pseudo force was acting on block.
$$F=ma+mg$$ in downward direction.
Spring Force =$$Kx$$ in upward direction
previous mean position be $$x_o$$
$$F_{net}=0$$
$$ma+mg=Kx_o$$
$$x_o=\dfrac{ma+mg}{K}$$
$$x_o=\dfrac{2\times 10+2\times 10}{800}$$
$$x_o=0.050 m=5 cm$$
This is extreme position now.
now, pseudo force stop acting , new mean position be $$x'$$
$$F_{net}=0$$
$$mg=Kx'$$
$$x'=\dfrac{mg}{K}$$
$$x'=\dfrac{2\times 10}{800}$$
$$x'=0.025 m=2.5 cm$$
so initially block moves from $$x=5 cm$$ to $$x=2.5 cm$$,
i.e block moves upwards
and it is given that downward direction is positive.
which means block is moving in negative direction,
so phase difference would be $$-\pi/2$$ i.e $$3\pi/2$$
Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite direction each time their displacement us half their amplitude. Their phase difference is :
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$$\dfrac{5\pi}{6}$$
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$$\dfrac{4\pi}{3}$$
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$$\dfrac{\pi}{6}$$
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$$\dfrac{2\pi}{3}$$
A standing wave arises on a string when two waves of equal amplitude, frequency, and wavelength traveling in opposite directions superimpose. If the frequency of two-component wäves is doubled, then the frequency of oscillation of the standing waves
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gets doubled
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gets halved
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remains unchanged
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changes but not by a factor of 2 or $$ 1 / 2 $$
Equations of a stationary and travelling waves are as follows $$y_1 = \sin {kx} \cos {\omega t}$$ and $$y_2 = a \sin {(\omega t - kx)} $$. The phase difference between two points $$x_1 = \pi/3k$$ and $$x_2 = 3 \pi/2K$$ is $$\phi _1$$ in the standing wave $$(y_1)$$ and is $$\phi _2$$ in travelling wave $$(y_2)$$ then ratio $$\phi _1/\phi _2$$ is
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1
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5/6
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3/4
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6/7
Two sine waves of slightly different frequencies $$f_1$$ and $$f_2 (f_1 > f_2)$$ with zero phase difference, same amplitudes, travelling in the same direction superimpose.
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Phenomenon of beats is always observed by human ear.
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Intensity of resultant wave is a constant.
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Intensity of resultant wave varies periodically with time with maximum intensity $$4 a^{2}$$ and minimum intensity zero.
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A maxima appears at a time $$1/[2(f_1 - f_2)]$$ later (or earlier) than a minima appears.
Figure shows the standing wave pattern at t = 0 due to superposition of waves given by $$y_1\space and\space y_2$$. N is anode and A an antinode. At this instant say t = 0, instantaneous velocity of points on the string
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is different fro different points
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is zero for all points
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changes with position of the point
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is constant but not equal to zero for all points
Explanation
$$y = y_1 + y_2 = (12 sin 5x) cos 4t$$
Maximum value of y-position in SHM of an element of the string that is located at an antinode = $$\pm 12 cm\space (sin 5x = pm 1)$$
For the position nodes amplitude should be xoro.
So, Sin 5x = 0 $$\Rightarrow 5x = n\pi$$
$$ x = \dfrac{n\pi}{5}$$
Where n = 0, 1, 2, 3,......
Value of amplitude at x = 1.8 cm
A = 12 sin (5 X 1.8) = 4.9 cm
At any instant say t = 0, instantaneous velocity of points on the string is zero for all points as at extreme position velocities of particles are zero.
The following equations represent progressive transverse waves
$$z_1 = A \cos (\omega t - kx)$$
$$z_2 = A \cos (\omega t + kx)$$
$$z_3 = A \cos (\omega t + ky)$$
$$z_4 = A \cos (2\omega t - 2ky)$$
A stationary wave will be formed by superposing
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$$z_1$$ and $$z_2$$
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$$z_1$$ and $$z_4$$
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$$z_2$$ and $$z_3$$
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$$z_3$$ and $$z_4$$
Explanation
a. The direction of wave must be opposite and frequencies will be same then by susperposition, standing wave formation takes place.
Amplitude of simple harmonic motion of a point on the string that is located at x = 1.8 cm will be
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3.3 cm
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6.7 cm
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4.9 cm
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2.6 cm
Explanation
$$y = y_1 + y_2 = (12 sin 5x) cos 4t$$
Maximum value of y-position in SHM of an element of the string that is located at an antinode = $$\pm 12 cm\space (sin 5x = pm 1)$$
For the position nodes amplitude should be xoro.
So, Sin 5x = 0 $$\Rightarrow 5x = n\pi$$
$$ x = \dfrac{n\pi}{5}$$
Where n = 0, 1, 2, 3,......
Value of amplitude at x = 1.8 cm
A = 12 sin (5 X 1.8) = 4.9 cm
At any instant say t = 0, instantaneous velocity of points on the string is zero for all points as at extreme position velocities of particles are zero.
Maximum value of the y-position coordinate in the simple harmonic motion of an element of the string that is located at an antinode will be
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$$\pm 6 cm$$
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$$\pm 8 cm$$
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$$\pm 12 cm$$
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$$\pm 3 cm$$
Two speakers are placed as shown in fig.
Mark out the correct statement(s)
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If a person is moving along AB, he will hear the sound as loud, faint, loud and so on
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If a person moves along CD, he will hear loud, faint, loud and so on
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If a person moves along AB,he will hear uniform intense sound
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If a person moves along CD, he will hear uniform intense sound
Explanation
At any point on line AB, the phase difference between two waves is zero and hence waves will interfere constructively.
Along CD, the phase difference changes and waves interfere constructively and destructively and, hence sound will be loud, faint and so on.
As a wave propagates
[IIT-JEE 1999]
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The wave intensity remains constant for a plane wave
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The wave intensity decreases as the inverse of the distance from the source for a spherical wave
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The wave intensity decreases as the inverse square of the distance from the source for a spherical wave
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The wave intensity of the spherical wave over the spherical surface centered at the source remains constant at all times
Explanation
For a travelling wave, the intensity of wave remains constant if it is a plane wave.
Intensity of wave is inversely proportional to the square of the distance from the source if the wave is spherical
$$\left (I = \frac{P} {4 \pi r^{2}} \right )$$
Intensity of spherical wave on the spherical surface centred at source always remains same. Here total intensity means power P.
Equation of motion in the same direction is given by
$$y_1=Asin(\omega t-kx), y_2 = A sin(\omega t-kx-\theta)$$ The amplitude of the medium particle will be
[BHU 2003]
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$$ 2A cos \dfrac{\theta}{2}$$
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$$2A cos \theta$$
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$$\sqrt{2} A cos \dfrac{\theta}{2}$$
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$$1.2f, 1.2\lambda$$
The superposing waves are represented by the following equations :$$y_1 = 5 sin 2 \pi (10 t - 0.1 x). y_2 = 10 sin 2\pi (20 t - 0.2x)$$ Ratio of intensities $$\dfrac{I_{(max)}}{I_{(MIN)}}$$ will be
[AIIMS 1995; KCET 2001]
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1
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9
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4
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16
A transverse sinusoidal wave of a amplitude a, wavelength $$\lambda$$ and frequency n is travelling on a stretched string. The maximum speed of any point on the string is v/10, where v is the speed of propagation of the wave. If $$a = 10^{-3} m $$ and $$v = 10 ms^{-1} m $$, then $$\lambda$$ and n are given by
[IIT 1998]
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$$\lambda = 2\pi \times 10^{-2}$$ m
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$$\lambda = 10^{-3}$$ m
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$$\lambda = \frac{10^{3}} {2\pi} $$ Hz
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$$ n = 10^{4}$$ Hz
Explanation
$$v_{max} = a \omega = \frac{v} {10} = \frac{10} {10} = m/sec$$
$$\Rightarrow a \omega = a \times 2 \pi n = 1 \Rightarrow n = \frac{10^{3}} {2 \pi}$$ ($$\therefore a = 10^{-3}$$ m)
Since $$ v = n \lambda \Rightarrow \lambda = \frac{v} {n} = \frac{10} {10^{3}/2 \pi} = 2 \pi \times 10^{-2}$$ m
Two waves having equations
$$x_1 = a sin (\omega t + \phi_1), x_2 = a sin(\omega t + \phi_2)$$
If in the resultant wave the frequency and amplitude remain equal to those of superimposing waves. Then phase difference between them is
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$$\dfrac{\pi}{6}$$
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$$\dfrac{2\pi}{3}$$
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$$\dfrac{\pi}{4}$$
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$$\dfrac{\pi}{3}$$
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Practice Class 11 Engineering Physics Quiz Questions and Answers
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