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CBSE Questions for Class 11 Engineering Physics Waves Quiz 4 - MCQExams.com
CBSE
Class 11 Engineering Physics
Waves
Quiz 4
Statement-1 : Where two vibrating tuning forks having frequencies $$256 Hz$$ and $$512 Hz$$ are held near each other, beats cannot be heard.
Statement-2 : The principle of superposition is valid only if the frequencies of the oscillations are nearly equal.
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Both the statements are true and statement-2 is the correct explanation of statement-1
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Both the statements are true but statement-2 is not the correct explanation of statement-1
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Statement-1 is true and statement-2 is false
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Statement-1 is false and statement-2 is true
Explanation
For beats to be heard the two frequencies should be close to each other. When two frequencies are $$f$$ and $$2f$$, the beat frequency is $$2f-f=f$$ which is the original frequency.
Thus no beats are heard. Principle of super position is valid for all frequencies.
Two waves of equal amplitude $$x_{o}$$ and equal frequency travel in the same direction in a medium. The amplitude of the resultant wave is :
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$$0$$
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$$x_{o}$$
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$$2x_{o}$$
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Between $$0$$ and $$2x_{o}$$
Explanation
Two waves of equal amplitude $$x_o$$ and equal frequency travel in the same direction in a medium.
$$\therefore$$ Amplitude of the resultant wave is given by,
$$R=x_1+x_2$$
$$R$$ is minimum when both are zero.
$$R$$ is maximum when both are $$x_o$$.
$$\therefore R_{max}=x_o+x_o=2x_o$$
$$\therefore $$ Amplitude of the resultant wave is given between $$0$$ and $$2x_o$$
The equation $$y=4+2\sin(6t-3x)$$ represents a wave motion with
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amplitude $$6$$ units
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amplitude $$2$$ units
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wave speed $$2$$ units
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wave speed $$1/2$$ units
Explanation
The sinusoidal part has maximum value of $$2$$. Thus $$2$$ units is amplitude.
$$y=y_{1}+A \sin(wt-kx)$$
$$v=\dfrac{w}{k}=\dfrac{6}{3}=2 m/s$$
Wave speed is $$2$$ units.
A wave pulse on a string has the dimension shown in figure. The wave speed is $$\nu= 1 \ cm/ s$$. If point O is a free end, the shape of wave at time $$t = 3s$$ is :
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0%
0%
0%
Explanation
In three seconds, the distance travelled by wave=$$1cm/s\times 3s=3cm$$
Hence the center of the wave, the crest, reaches the point O' in the time.
The amplitude of the wave becomes $$1cm+1cm=2cm$$
Hence correct answer is option D.
Two sound waves, each of amplitude $$A$$ and frequency $$\omega$$, superpose at a point with a phase difference of $$\displaystyle \dfrac{\pi}{2}$$. The amplitude and the frequency of the resultant wave are, respectively
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$$A^2,\omega$$
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$$2A,\omega$$
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$$A,\omega$$
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$$\sqrt{2}A,\omega$$
Explanation
Let the two waves be $$A\ \sin\ wt,\ A\ \cos\ wt$$
$$\therefore y=y_{1}+y_{2}=\ A\ \sin\ wt+A\ \cos\ wt$$
$$=\sqrt{2}A\left ( \dfrac{1}{\sqrt{2}}\sin\ wt+\ \cos\ wt\dfrac{1}{\sqrt{2}} \right )$$
$$=\sqrt{2}A\ \ \ \sin \left ( wt+\dfrac{\pi}{4} \right )$$
amplitude is $$\sqrt{2}A$$, frequency is same, i.e. $$w$$
A wave represented by $$y=100\sin(ax+bt)$$ is reflected from a dense plane at the origin. If $$36\%$$ of energy is lost and rest of the energy is reflected, then the equation of the reflected wave will be
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$$ y=-8.1\sin ({\it ax-bt} )$$
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$$y=8.1\sin(ax+bt)$$
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$$ y=-80\sin (ax-bt )$$
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$$ y=-10\sin (ax-bt)$$
Explanation
Given, $$36\%$$ of energy is lost, energy of wave reflected back is $$64\%$$
$$I \alpha A^{2} \Rightarrow \ \ \ \ A_{2}=\dfrac{8}{10}\times A_{1}$$
$$\therefore \dfrac{8}{10}\times 100=80$$
reflected wave will be in opposite direction as it's reflected from a dense plane.
$$\Rightarrow y=-80\ \sin\ (ax-bt)$$
A plane progressive wave of frequency $$25 Hz$$, amplitude $$2.5 \times 10^{-5}m$$ and initial phase zero moves along the negative x-direction with a velocity of $$300 ms^{-1}$$ . A and B are two points $$6m$$ apart on the line of propagation of the wave. At any instant the phase difference between A and B is $$\Theta$$ . The maximum difference the displacement of the particles at A and B is $$\Delta$$ , then
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$$\Theta =\pi$$
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$$\Theta =0$$
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$$\Delta=0$$
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$$\Delta=5\times 10^{-5}m$$
Explanation
Given : $$f=25Hz , A =2.5\times10^{-5}m , v=300m/s$$,
from , $$v=f\lambda$$,
or $$\lambda=v/f=300/25=12m$$,
As points A and B are 6cm apart hence path difference between A and B will be , $$\Delta x=6cm$$,
by $$\Delta \phi=\dfrac{2\pi}{\lambda}\Delta x$$,
$$\Delta \phi=\dfrac{2\pi}{12}\times6=\pi$$ (phase difference between A and B),
Now, displacement for particle A,
$$y_{A}=A\sin\omega t$$,
or $$y_{A}=2.5\times10^{-5}\sin\omega t$$,
and ,
displacement for particle B,
$$y_{B}=A\sin(\omega t+\pi)$$, (as wave is travelling in -ive x-direction)
or $$y_{B}=2.5\times10^{-5}\sin(\omega t+\pi)$$,
or
$$y_{B}=-2.5\times10^{-5}\sin\omega t$$,
therefore , difference in displacements of A and B,
$$y_{A}-y_{B}=[2.5\times10^{-5}-(-2.5\times10^{-5})]\sin\omega t$$,
this difference will be maximum when , $$\sin\omega t=1(maximum)$$,
therefore , $$\Delta$$
=$$[2.5\times10^{-5}-(-2.5\times10^{-5})]=5\times10^{-5}m$$
A $$40\ cm$$ long brass rod is dropped, one end first on to a hard floor but it is caught before it topples over. With an oscilloscope it is determined that the impact produces a $$3\ kHz$$ tone. The speed of sound in brass is:
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$$1200\ m/s$$
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$$2400\ m/s$$
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$$3600\ m/s$$
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$$3000\ m/s$$
Explanation
Both ends are free and therefore antinodes are formed.
The relation between the wavelength of the wave and the length of the rod for fundamental frequency will be:
$$\Rightarrow l=\dfrac{\lambda}{2} \ \ \Rightarrow \lambda=2l$$
The speed of the wave in the rod is:
$$v=f\lambda = 2\times 40\times 3\times 10^{3}$$
$$=2400\ ms^{-1}$$
When a sound wave of wavelength $$\lambda$$ is propagating in a medium the maximum velocity of the particle is equal to wave velocity. The amplitude of the wave is
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$$\lambda$$
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$${\dfrac{\lambda}{2}}$$
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$$\displaystyle \dfrac{\lambda}{2\pi}$$
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$$\displaystyle \dfrac{\lambda}{4\pi}$$
Explanation
wave velocity of particle $$v=\dfrac{w}{K}$$
maximum particle velocity $$v_{2}=Aw$$
given $$Aw=\dfrac{w}{K}\ \ \ \ \ \ \Rightarrow A=\dfrac{1}{K}=\dfrac{1}{\dfrac{2\pi}{\lambda}}=\dfrac{\lambda}{2\pi}$$
A wave travelling in positive X-direction with A = 0.2 m velocity = 360 m/s and $$\lambda$$= 60 m, then correct expression for the wave is : -
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y = 0.2 sin $$\left [ 2\pi (6t+\frac{X}{60}) \right ]$$
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y = 0.2 sin $$\left [\pi (6t+\frac{X}{60}) \right ]$$
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y = 0.2 sin $$\left [ 2\pi (6t-\frac{X}{60}) \right ]$$
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y = 0.2 sin $$\left [\pi (6t-\frac{X}{60}) \right ]$$
A particle starts from a point $$P$$ at a distance of $$A/2$$ from the mean position $$O$$ & travels towards left as shown in the figure. If the time period of SHM, executed about $$O$$ is $$T$$ and amplitude $$A$$ then the equation of motion of particle is :
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$$x = A \sin\left ( \dfrac{2\pi }{T} t+\dfrac{\pi }{6}\right )$$
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$$x = A \sin\left ( \dfrac{2\pi }{T} t+\dfrac{5\pi }{6}\right )$$
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$$x = A \cos\left ( \dfrac{2\pi }{T} t+\dfrac{\pi }{6}\right )$$
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$$x = A \cos \left ( \dfrac{2\pi }{T} t+\dfrac{\pi }{3}\right )$$
Explanation
$$\begin{array}{l} x=A\cos \, (\omega t+\theta ) \\ and\, \, \, v=\, \, -\, \omega A\sin \, \, (\omega t+\theta ) \\ \, \, at\, \, \, \, \, t=0, \\ \, x=\, A\, cos\theta \, \, \, \, \, and\, \, \, \, \, v=\, -\, \omega A\sin \theta \\ \, \, \, \, \, \therefore \, \, \, \, \dfrac { A }{ 2 } =A\, cos\theta \\ \Rightarrow \theta ={ \cos ^{ -1 } }\left( { \dfrac { 1 }{ 2 } } \right) =\dfrac { \pi }{ 3 } \\ now, \\ \, \, \, \, \, \, x=A\cos \, (\omega t+\frac { \pi }{ 3 } ) \\ \, \, \, \, \, \, \, \, \, \, =A\cos \, (\dfrac { { 2\pi } }{ T } .t+\dfrac { \pi }{ 3 } ) \\ \, \, \, \, \, \, \, \, \, \, =Asin\, (\dfrac { { 2\pi } }{ T } .t+\dfrac { \pi }{ 2 } +\dfrac { \pi }{ 3 } ) \\ \, \, \, \, \, \therefore \, \, \, A\, sin\, (\dfrac { { 2\pi } }{ T } .t+\dfrac { { 5\pi } }{ 6 } ) \\ so\, the\, correct\, option\, is\, \, B. \end{array}$$
The displacement of a particle varies according to the relation $$x = 3 \sin 100t + 8 \cos ^{2} 50t$$ . Which of the following is/are correct about this motion .
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the motion of the particle is not S.H.M.
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the amplitude of the S.H.M. of the particle is $$5$$ units
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the amplitude of the resultant S.H. M. is $$\sqrt{73}$$ units
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the maximum displacement of the particle from the origin is $$9$$ units .
Explanation
$$x=3\sin100t+8\cos^2 50t$$
$$x=3\sin100t+4(1-cos 100t) $$
$$x=3\sin100t-4\cos 100t+4 $$
$$x=3\sin100t-4\cos 100t+4 $$
Maximum value of$$(3\sin100t-4\cos 100t)$$ is 5
Amplitude=5
the maximum displacement of the particle from the origin is $$5+4$$ = 9
Hence, Option B and D is answer.
A small mass executes linear SHM about O with amplitude a and period T. Its displacement from O attime T/8 after passing through O is:
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$$\dfrac{a}{8}$$
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$$\dfrac{a}{2\sqrt{2}}$$
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$$\dfrac{a}{2}$$
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$$\dfrac{a}{\sqrt2}$$
Explanation
Using the formula $$x=a\times cos(wt)$$ for SHM, and substituting $$t=\dfrac{T}{8 }$$and$$ w=\dfrac{w\times pi}{T}$$, we get $$x=a\times cos(\dfrac{pi}{4})=\dfrac{a}{ \sqrt(2)}$$
The amplitude of the vibrating particle due to superposition of two $$SHMs$$,
$$y_{1}= \sin \left ( \omega t+\dfrac{\pi }{3} \right )$$ and $$y=2 \sin \omega t$$ is:
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$$1$$
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$$\sqrt{2}$$
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$$\sqrt{7}$$
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$$2$$
Explanation
$$\begin{array}{l} { A^{ 2 } }=A_{ 1 }^{ 2 }+A_{ 2 }^{ 2 }+2{ A_{ 1 } }{ A_{ 2 } }\cos \emptyset \\ { A^{ 2 } }={ 1^{ 2 } }+{ 2^{ 2 } }+2\times 1\times 2\cos \left( { \pi /3 } \right) \\ { A^{ 2 } }=1+4+4\times \frac { 1 }{ 2 } \\ { A^{ 2 } }=5+2 \\ A=\sqrt { 7 } \end{array}$$
A cylindrical block of density $$\rho $$ is partially immersed in a liquid of density $$3\ \rho$$. The plane surface of the block remains parallel to the surface of the liquid. The height of the block is $$60\ cm$$. The block performs SHM when displaced from its mean position. [Use $$g = 9.8 m/s^{2}$$]
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the maximum amplitude is $$20\ cm$$.
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the maximum amplitude is $$40\ cm$$
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the time period will be $$\dfrac{2\pi}{7}$$ seconds.
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$$none$$
Explanation
When cylindrical block is partially immersed
$$F_B=mg⇒3ρAyg=ρA(60 \times 10^{−2})g$$
$$y=20cm$$
$$⇒ Maximum \, amplitude =20cm$$
Restoring force when it is slightly depressed by an amount of x.
$$F=−(ΔVσg)=−(Aσg)x$$
$$T$$= $$2 \pi \sqrt{\dfrac{m}{Aσg}}$$ = $$2\pi \sqrt{\dfrac{ρAh}{3Aρg}} $$ = $$2 \pi \sqrt{\dfrac{h}{3g}} $$
$$= 2 \pi \sqrt{\dfrac{60 \times 10^{-2}}{3 \times 9.8}} = \dfrac{2 \pi}{7}$$
Assertion :If Amplitude of SHM is doubled, the periodicity wall remain same.
Reason :Amplitude and periodicity are two independent characteristics of SHM.
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Assertion & Reason are true and the reason explains the assertion
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Assertion & Reason are true and the reason does not explain the assertion
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Assertion is true. Reason is false
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Both assertion and reason are false
Explanation
Periodicity depends on other characteristics like length in case of pendulum, mass in case of spring mass system. But it is independent of amplitude.
Statement-1 : Superposition principle is applicable only for small disturbances.
Statement-2 : Superposition principle is applicable only for non-linear waves.
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If both the statements are true and statement-2 is the correct explanation of statement-1
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If both the statements are true but statement-2 is not the correct explanation of statement-1
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If statement-1 is true and statement-2 is false
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If statement-1 is false and statement-2 is true
Explanation
Superposition principle, states that, for all linear systems, the net response at a given place and time caused by two or more stimuli is the sum of the responses which would have been caused by each stimulus individually.
A block is placed on a horizontal plank. The plank is performing $$SHM$$ along a vertical line with an amplitude of $$40\ cm$$. The block just loses contact with the plank when the plank is momentarily at rest. Then:
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The period of its oscillations is $$2\pi /5 sec$$.
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The block weights on the plank double its
weight
when the plank is at one of the positions of
momentary rest.
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The block weights $$1.5$$ times its weight on the plank halfway down from the mean position.
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T
he block weights its true weight on the plank when the velocity of the plank is maximum.
Explanation
Given:
$$A = 40\,cm = 0.4\,m$$
Since the block loses contact $$(N = 0)$$ when the plank is momentarily at rest at the top. Then the force balance equation on the block at the top is:
$$N-mg = ma_{max}$$
$$\Rightarrow a_{max} = -g = -10\,m/s^2$$
But
$$a = -\omega^2 x$$, then at the top:
$$a_{max} = -\omega^2 A$$
$$\Rightarrow \omega = \sqrt{-\dfrac{a_{max}}{A}} = \sqrt{\dfrac{10\,m/s^2}{0.4}}$$
$$\Rightarrow \omega = 5\, rad/s$$
(A) Thus, $$T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{5}\,s$$
(B) Now, at the bottom-most point, $$(x = -A)$$:
$$a = -\omega^2x = \omega^2A = g$$
Then the force balance equation on the block:
$$N - mg = ma$$
$$\Rightarrow N = m(g + a) = 2mg$$
Thus, t
he weight of block is doubled at the bottom point.
(C) At the halfway down position, $$(x = \dfrac{A}{2})$$:
$$a = -\omega^2x = -\omega^2\dfrac{A}{2} = -\dfrac{g}{2}$$
Then the force balance equation on the block:
$$N - mg = ma$$
$$\Rightarrow N = m(g + a) = \dfrac{1}{2}mg$$
Thus, t
he weight of block is halved at the halfway down point.
(D) The velocity of plank is maximum at zero position $$(x = 0)$$:
$$a = -\omega^2x = 0$$
Then the force balance equation on the block:
$$N - mg = ma$$
$$\Rightarrow N = m(g + a) = mg$$
Thus, t
he weight of block remains the same
.
When a wave pulse travelling in a string is reflected from a rigid wall to which string is tied as shown in figure. For this situation two statements are given below.
(1) The reflected pulse will be in same orientation of incident pulse due to a phase change of $$\pi$$ radians
(2) During reflection the wall exert a force on string in upward direction
For the above given two statements choose the correct option given below.
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Only (1) is true
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Only (2) is true
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Both are true
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Both are wrong
Explanation
Reflected pulse will be inverted as it is reflected by a denser medium.
Since there will be phase change of $$\pi$$.
The wall exerts force in downward direction.
The type of waves is/ are
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Progressive
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Stationary
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Both (a) and (b)
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None of these
If the vibration of particles in a wave is perpendicular to the direction of propagation of wave, then it is called,
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Transverse waves
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Reflective waves
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Incident waves
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None of these
Reflection of a light wave at a fixed point results in a phase difference between incident and reflected wave of
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$$\dfrac {3\pi}{2}$$
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$$2\pi $$
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$$\pi$$
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$$\dfrac {\pi}{2}$$
Explanation
Reflections of light at the interface between media often produce phase differences. The phase difference between incident and reflected wave is $$2\pi$$ and $$\pi / 2$$ radians.
Two small boats are $$10m$$ apart on a lake. Each pops up and down with a period of $$4.0\ \text{seconds}$$ due to wave motion on the surface of the water. When one boat is at its highest point, the other boat is at its lowest point. Both boats are always within a single cycle of the waves. The speed of the waves is
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$$2.5\ \text{m/s}$$
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$$5.0\ \text{m/s}$$
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$$14\ \text{m/s}$$
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$$40\ \text{m/s}$$
Explanation
$$\dfrac{\lambda}{2}=10\ \text{m}\Rightarrow \lambda=20\ \text{m}$$
$$\text{Time period}= T=4\ \text{sec}$$
$$ V=\dfrac{\lambda}{T}=\dfrac{20}{4}=5\ \text{m/s}$$
If the frequency of a wave is increased by 25 %, then the change in its wavelength will be:
(medium not changed)
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20 % increase
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20 % decrease
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25 % increase
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25 % decrease
Explanation
Since, the medium has not changed, speed of wave remains
same.
$$v=\nu_1 \lambda_1=\nu_2 \lambda_2$$
$$\nu_2=1.25 \nu_1$$
$$\lambda_2=\dfrac{\nu_1}{\nu_2} \lambda_1=\dfrac{1}{1.25} \lambda_1=(4/5) \lambda_1$$
$$ \dfrac {\Delta \lambda}{\lambda _1} = \dfrac{\lambda_2- \lambda_1}{\lambda_1} \times 100= \dfrac{(4/5)\lambda_1- \lambda_1}{\lambda_1} \times 100=-(1/5)\times 100=-20$$ %
option "B" is correct.
An object is vibrating at $$50$$ hertz. What is its time period?
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$$0.02$$ s
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$$0.2$$ s
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$$2$$ s
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$$20.0$$ s
Explanation
From the definition of frequency $$\nu= \dfrac{1}{T}$$
and we are given $$\nu=50 hertz $$.
Substituting, we have $$50 = \dfrac{1}{T}$$ or
$$T = 0.02 s$$
Snapshot for a rope shown, at an instant is carrying a travelling wave towards right, created by source vibrating with the frequency "n". Then,
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the speed of the wave is (4 n) (distance ab)
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the particle at point a will be in the present phase of d after $$\frac{4}{3n}$$ sec.
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the phase difference between b and e is $$\frac{3\pi}{2}$$
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the wave is harmonic
Explanation
As it is a sinusoidal curve so wave is obviously harmonic,
Now $$V = \gamma \lambda, $$ where $$\gamma$$ is frequency
$$= n(4ab)$$
$$= 4n(distance\;ab)$$
Phase delay of $$d= \dfrac{3\lambda /4}{v} = \dfrac{\dfrac{3}{4}(4ab)}{4n(ab)} = \dfrac{3}{4n}$$
Phase difference between b and e $$= 3\times\dfrac{\pi}{2} = \dfrac{3\pi}{2}$$
The frequency of a man's voice is 300 Hz and its wavelength is 1 meter. If the wavelength of a child's voice is 1.5 m, then the frequency of the child's voice is :
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200 Hz
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150 Hz
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100 Hz
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350 Hz.
Explanation
$$\nu_1\lambda_1= \nu_1\lambda_1$$ since $$v= \nu\lambda$$ is same for both a man and child.
$$ \therefore 300 \times 1 = \nu_2 \times 1.5$$
$$ \Rightarrow \nu_2 = 200 \: Hz$$
Sound waves are .................... waves.
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Transverse
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Longitudinal
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Straight Line
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None of these
Explanation
Sound waves are called longitudinal waves because sound requires a material medium for their propagation.
Hence, the correct option is B
Frequency of the sinusoidal wave, y = 0.40 cos (2000t + 0.080) would be :
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100 at Hz
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2000 Hz
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20 Hz
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318Hz
Explanation
The general form of a sinusoidal wave is,
$$y=A cos(2\pi\nu t+\phi)$$, where,
$$A=$$amplitude of the wave
$$\nu= $$frequency
$$\phi=$$phase shift
Comparing this with the given equation then the frequency will comes out to be,
$$\omega = 2\pi f $$
Here $$\omega = 2000$$ Therefore will get frequency as,
$$\omega = 2\pi f = 2000$$
Calculating,
$$f=318\ Hz$$
The frequency of sound waves is 11 kHz and its wavelength is 20 cm, then the velocity of sound waves is :
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220 m s$$^{-1}$$
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220 cm s$$^{-1}$$
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2200 cm s$$^{-1}$$
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2200 m s$$^{-1}$$
Explanation
Velocity of sound wave $$V=n \lambda$$ i.e. the product of the frequency and amplitude of the wave.
So, $$V=11 \times 10^3 \times 20 \times 10^{-2}=2200$$ m/s
An ultrasonic source emits sound of frequency 220 kHz in air. If this sound meets a water surface, what is the wavelength of the transmitted sound? (At the atmospheric temperature, speed of sound in air $$=352 m s^{-1} and in water = 1.496 m s^{-1}$$)
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$$5.8 \times 10^{-3} m$$
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$$6.8 \times 10^{-3} m$$
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$$7.8 \times 10^{-3} m$$
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$$8.8 \times 10^{-3} m$$
Explanation
Here, $$v=220 kHz = 220 \times 10^3$$
$$=2.2 \times 10^5 Hz$$;
Speed of sound in air, $$v_a = 352 m s^{-1}$$
Speed of sound in water, $$v_w = 1.496 m s^{-1}$$
The transmitted sound : The transmitted ultrasonic sound travels in water. If $$\lambda_w$$ is wavelength of ultrasonic sound in water, then
$$\lambda_w = \displaystyle \frac{v_w}{v} = \frac{1,496}{2.2 \times 10^5} = 6.8 \times 10^{-3} m$$
Which of the following statements is correct?
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Both, sound and light waves in air are longitudinal.
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Both, sound and light waves in air are transverse.
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Sound waves in air are transverse and light waves are longitudinal.
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Sound waves in air are longitudinal and light waves are transverse.
Explanation
sound waves are longitudinal waves and light being a EM wave is transverse in nature .
so the answer is D.
Elastic waves in solid are :
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Transverse Only
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Longitudinal Only
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Either transverse or longitudinal
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Neither transverse nor longitudinal
Explanation
Elastic wave
, motion in a medium in which, when particles are displaced, a force proportional to the displacement acts on the particles to restore them to their original position. so depending on the forces acting the waves may be either longitudinal or transverse .
the answer is C.
An oscilloscope is basically designed to convert .................... .
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Visual signals to electrical signals
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Sound signals to electrical signals
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Electrical signals to visual signals
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Sound signals to visual signals
Explanation
Solution
Oscilloscope helps us to display and test variations in voltage signals with time in graph form.
Therefore it converts electrical signal to visual signal.
The correct option is C
In a stationary wave,
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Phase of all vibrating particles is same.
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Phase of different particles is different.
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Phase of particles between the two nodes is same.
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None of these
Explanation
Solution
A stationary wave is produced when two waves of same wavelength and amplitude but moving in opposite direction superimpose.
Particles at node are stationary. particles at antinodes move with high velocity
All those particles between two nodes vibrates with same phase
The correct option is C
Elastic waves need material medium for their propagation.
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True
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False
Explanation
yes , the elastic waves needs material medium for their propagation
so the answer is A.
Consider a function $$y = 10 \sin^{2} (100\pi t + 5 \pi z)$$ where $$y, z$$ are in $$cm$$ and $$t$$ is $$second. $$
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the function represents a travelling, periodic wave propagating in (-z) direction with speed $$20 m/s.$$
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the function does not represent a travelling wave.
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the amplitude of the wave in $$5 cm.$$
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the amplitude of the wave is $$10 cm.$$
Explanation
$$\sin^2\theta = \dfrac{1}{2}(1-\cos(2\theta))$$.
So,
$$y=5[1-\cos(200\pi t+10 \pi z)]$$.
The above equation represents a plane progressive wave along -z axis with amplitude $$5 cm$$.
$$\omega = 200 \pi$$ and $$k=10 \pi$$
So velocity $$v = \dfrac{\omega}{k}=20cm/s$$
A string 1m long is drawn by a 300Hz vibrator attached to its end. The string vibrates in 3 segments. The speed of transverse waves in the string is equal to
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100 m/s
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200 m/s
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300 m/s
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400 m/s
Explanation
Wavelength, $$ \lambda $$ $$= 2L$$.
Where, $$L$$ is the length of the segment or number of segments.
Velocity, $$v =$$ $$ \lambda f$$ $$=300\times \dfrac{2}{3} = 200\ m/s $$
769Hz longitudinal wave in air has a speed of 344m/s.
At a particular instant, what is the phase difference (in degrees) between two points 5.0 cm apart?
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30
0%
40
0%
45
0%
60
Explanation
Given that,
The frequency of wave is $$\nu = 769 Hz$$
The speed of wave is $$v = 344 ms^{-1}$$
The wavelength of wave is given by
$$\lambda = \dfrac{v}{\nu}$$
$$\lambda = \dfrac{344}{769}$$
$$\lambda = 0.447 m = 44.7 cm$$
Hence,
the phase difference between two points 5.0 cm apart is
$$\dfrac{5}{44.7} \times 360^\circ = 0.1118 \times 360^\circ = 40.24^\circ \approx40^\circ$$
A sound wave travels with a speed of $$330 ms^{-1}$$ in air. If the wavelength of the wave is 3.3m, then the frequency of the wave is ______.
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50 Hz
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100 Hz
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150 Hz
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200 Hz
Explanation
since $$ v= f \times \lambda$$
$$ 330= f \times 3.3 $$
$$ f = 100$$ Hz
so the answer is B.
A composition string is made up by joining two strings of different masses per unit length $$\longrightarrow \mu $$ and $$4\mu.$$ The composite string is under the same tension. A transverse wave pulse : $$Y=(6 mm) \sin (5t+40x)$$, where '$$t$$' is in seconds and '$$x$$' in meters, is sent along the lighter string towards the joint. The joint is at $$x=0.$$ The equation of the wave pulse reflected from the joint is
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$$(2 mm) \sin(5t-40x)$$
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$$(4 mm) \sin(40x-5t)$$
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$$- (2 mm) \sin(5t-40x)$$
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$$(2 mm) \sin(5t-10x)$$
Explanation
Since the wave is travelling from low to more dense medium, thus there will be inversion of the reflected wave . Since it travels back from origin, thus $$40x$$ will become $$-40x.$$ Also, amplitude of reflected wave is given by:
$${ A }_{ r }=\dfrac { Z_{ 1 }-{ Z }_{ 2 } }{ { Z }_{ 1 }+Z_{ 2 } } A.\quad Z=\mu c=\mu \sqrt { \dfrac { T }{ \mu } } =\sqrt { \mu T } \\ Thus\quad { A }_{ r }=\dfrac { \sqrt { \mu T } -\sqrt { 4\mu T } }{ \sqrt { \mu T } +\sqrt { 4\mu T } } A=\dfrac { -1 }{ 3 } A\\ $$
Thus it become $$-6/3=-2$$.
A pulse shown in the figure is reflected from the rigid wall A and then from free end B. The shape of the string after these 2 reflection will be :
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0%
0%
0%
Explanation
Explanation :
Phase change because of reflection from an inflexible limit $$(\mathrm{A})=\pi$$.
Phasechange because of reflection from a more uncommon limit $$(\mathrm{B})=0$$.
Total phase change due to two reflections $$=\pi$$.
Therefore, The correct option is 'A'
A wave is represented by the equation $$y=10 \sin 2\pi(100t-0.02x)+10 \sin 2\pi(100+0.02x)$$. The maximum amplitude and loop length are respectively
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$$20$$ units and $$30$$ units
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$$20$$ units and $$25$$ units
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$$30$$ units and $$20$$ units
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$$25$$ units and $$20$$ units
Explanation
$$x= 10\sin 2\pi \Big(100t-0.02x\Big)+10\sin 2\pi \Big(100t+0.02x\Big)$$.
$$\Rightarrow 10 [\sin A+\sin B]$$,
where
$$B= 2\pi \Big(100t-0.02x\Big)$$ and
$$A= 2\pi \Big(100t+0.02x\Big)$$.
Thus,
$$\Rightarrow 10 [2\sin \dfrac{A+B}{2}\sin \dfrac{A-B}{2}]$$
$$\Rightarrow 20\sin (2\pi 100t) \sin (2\pi 0.02x)$$.
Comparing the above equation with standard standing wave equation, we get
amplitude $$= 20$$ and wave vector$$k=\dfrac{2\pi}{\lambda}=2\pi \times 0.02 \Rightarrow \lambda = 50 $$.
Therefore, the loop length = $$\dfrac{\lambda}{2}= 25$$
A particle is executing SHM of amplitude $$A$$ about the mean position $$x=0$$. Which of the following cannot be a possible phase difference between the positions of the particle at $$x=+{A}/{2}$$ and $$x=-{A}/{\sqrt{2}}$$.
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$$75^\circ$$
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$$165^\circ$$
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$$135^\circ$$
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$$195^\circ$$
Explanation
sine wave starts from zero,but for sin 90 degree its $$ \dfrac{1}{2}$$ whereas for sin 135 degree its $$ \dfrac{1}{ \sqrt{2} }$$ hence for sin 135 degree its value will be $$ \dfrac{1}{ \sqrt{2} }$$
therefore option (c) is correct
A longitudinal wave has a compression to compression distance of 10 m. It takes the wave 5 s to pass a point. Find the velocity of wave.
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2 m/s
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3 m/s
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4 m/s
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6 m/s
Explanation
The speed of wave is given by
$$v = \nu \lambda$$
where, v is the speed of the wave,$$\nu$$ is the frequency of wave and
$$\lambda$$ is its wavelength.
But, $$\nu = \dfrac{1}{T}$$
where T is a period of propagation of the wave.
Hence,
$$v = \dfrac{\lambda}{T}$$
$$v = \dfrac{10}{5} = 2 ms^{-1}$$
The resultant amplitude due to superposition of two waves $$y_{1} = 5 \sin (wt-kx)$$ and $$y_{2}=-5 \cos (wt-kx-150^{o})$$
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$$5$$
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$$5\sqrt{3}$$
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$$5\sqrt{2-\sqrt{3}}$$
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$$5\sqrt{2+\sqrt{3}}$$
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$$5\sqrt{3-\sqrt{2}}$$
Explanation
Given :
$${ y }_{ 1 }=5\sin(wt+kx)\\ { y }_{ 2 }=5\cos(wt+kx+{ 150 }^{ \circ })\\ y={ y }_{ 1 }+{ y }_{ 2 }$$
where $$y$$ is the resultant wave
The amplitude of this resultant wave is given by the formula:
$$ A = \sqrt { { (A }_{ 1 })^ 2+({ A }_{ 2 })^ 2+2{ A }_{ 1 }{ A }_{ 2 }\cos\phi } $$
Here;
$${ A }_{ 1 }=5$$
$${ A }_{ 2 }=5$$
$$\phi ={ 150 }^{ \circ }$$
on substituting the values on the given formula:
$$ A=\sqrt { (5)^ 2+(5)^ 2+2\times 5\times 5\times \cos{ 150 }^{ \circ } } $$
$$ =\sqrt { 25+25-2\times 25\times \frac { 1 }{ 2 } } .................(\cos{ 150 }^{ \circ }=-\dfrac { 1 }{ 2 } )$$
Hence
A=5 (option A is correct)
What happens when a sound wave is reflected from the boundary of a denser medium? The compression of the incident wave is returned as a
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rarefaction
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crest
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trough
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compression
Explanation
Sound wave goes through $$\pi$$ phase change when reflected from the boundary of a denser medium. Hence, the compression of the incident wave is returned as a compression.
The resultant amplitude, when two waves of same frequency but with amplitudes $$a_1$$ and $$a_2$$ superimpose with a phase difference of $$\pi/2$$ will be
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$$a_1^2 + a_2^2$$
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$$\sqrt{a_1^2 + a_2^2}$$
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$$a_1 - a_2$$
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$$a_1 + a_2$$
Explanation
The equation of the two superimposing waves can be give by
$$y_1=a_1\sin(\omega t)$$ and $$y_2=a_2\sin(\omega t +\pi/2)=a_2\cos(\omega t)$$.
So, the equation of the resultant wave is $$y=y_1+y_2=a_1\sin(\omega t)+a_2\cos(\omega t)$$.
Assuming $$a_1=a\cos(\phi)$$ and $$a_2=a\sin(\phi)$$, we get
$$y=a\sin(\omega t+\phi)$$.
Here $$a^2_1+a^2_2=a^2(\sin^2(\phi)+\cos^2(\phi))=a^2\Rightarrow a=\sqrt{a^2_1+a^2_2}$$.
Two waves of same amplitude and same frequency reach a point in a medium simultaneously. The phase difference between them for resultant amplitude to be zero, will be
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$$4\pi$$
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$$2\pi$$
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$$\pi$$
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$$0^{\small\circ}$$
Explanation
The equation of the two superimposing waves can be give by
$$y_1=a\sin(\omega t)$$ and $$y_2=a\sin(\omega t +\phi)$$. So, the equation of the resultant wave is $$y=y_1+y_2=a(\sin(\omega t)+\sin(\omega t+\phi))=0\Rightarrow \sin(\omega t)=-\sin(\omega t+\phi)\Rightarrow \phi=\pi, 3\pi...$$.
A particle performing SHM is found at its equilibrium at t=1sec, and it is found to have a speed of 0.25 m/s at t=2 sec. If the period of oscillation is 6 sec , calculate amplitude of oscillation.
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$$\displaystyle\frac{3}{2\pi}m$$
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$$\displaystyle\frac{3}{4\pi}m$$
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$$\displaystyle\frac{6}{\pi}m$$
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$$\displaystyle\frac{3}{8\pi}$$
Explanation
Since a SHM can be represented by $$x=A\sin { \left( \omega t+\phi \right) } $$
$$\Rightarrow x=A\sin { \left( \left( \dfrac { 2\pi }{ T } \right) t+\phi \right) } $$ ; $$T=6s$$
So the equation becomes: $$x=A\sin { \left( \dfrac { \pi t }{ 3 } +\phi \right) } $$
We are given that at $$t=1s, x=0 $$; so we have:
$$0=A\sin
{ \left( \dfrac { \pi }{ 3 } +\phi \right) } \\ \Rightarrow \left(
\dfrac { \pi }{ 3 } +\phi \right) =0\\ \Rightarrow \phi =-\dfrac { \pi
}{ 3 } \\ \Rightarrow x=A\sin { \left( \dfrac { \pi t }{ 3 } -\dfrac {
\pi }{ 3 } \right) } \\ \Rightarrow v=\dfrac { dx }{ dt } =\dfrac { d
}{ dt } \left( \sin { \left( \dfrac { \pi t }{ 3 } -\dfrac { \pi }{ 3 }
\right) } \right) =\dfrac { \pi A }{ 3 } \cos { \left( \dfrac { \pi t }{
3 } -\dfrac { \pi }{ 3 } \right) } $$
Now we are given that at $$t=2s, \left| v \right| =0.25m/s$$,
So we have:
$$0.25=\left|
\dfrac { \pi A }{ 3 } \cos { \left( \dfrac { 2\pi }{ 3 } -\dfrac { \pi
}{ 3 } \right) } \right| \\ \Rightarrow 0.25=\left| \dfrac { \pi A }{ 3
} \cos { \left( \dfrac { \pi }{ 3 } \right) } \right| \\ \Rightarrow
0.25=\dfrac { \pi A }{ 6 } \\ \Rightarrow A=\dfrac { 3 }{ 2\pi } $$
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Practice Class 11 Engineering Physics Quiz Questions and Answers
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