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Waves Test - 9

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Waves Test - 9
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  • Question 1
    1 / -0
    A transverse wave travels on a taut steel wire with a velocity of vv when the tension in it is 2.06×104N2.06\times { 10 }^{ 4 }N. When the tension is changed to TT, the velocity changed to v/2v/2. The value of TT is closet to:
    Solution
    Velocity of sound in a medium
    v=Tμv = \sqrt{\dfrac{T}{\mu}}

    vTv \propto \sqrt{T}

    v1v2=T1T2\Rightarrow \dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1}{T_2}}

    Given T1=2.06×104NT_1 = 2.06 \times 10^4N

    v2=v2v_2 = \dfrac{v}{2}

    v1=vv_1 = v

    vv/2=2.06×104T\dfrac{v}{v/2} = \sqrt{\dfrac{2.06\times 10^4}{T}}

    4=2.06×104T4 = \dfrac{2.06\times 10^4}{T}

    T=.515×104NT = .515 \times 10^4 N

    T=5.15×103NT = 5.15 \times 10^3N
  • Question 2
    1 / -0
    The displacement of a particle varies according to the relation x=4(cosπt+sinπt)x = 4(\cos \pi t + \sin \pi t). The amplitude of the particle is
    Solution
    When the displacement , x=acoswt+bsinwtx=a\cos wt+b\sin wt, the amplitude will be A=a2+b2A=\sqrt{a^2+b^2}
    Here, a=4,b=4a=4, b=4 so A=42+42=42A=\sqrt{4^2+4^2}=4\sqrt 2
  • Question 3
    1 / -0
    Amplitude of a wave is represented by A=ca+bcA = \dfrac {c}{a + b - c}. Then resonance will occur when
    Solution
    A=ca+bcA = \dfrac {c}{a + b - c}: when b=0,a=cb = 0, a = c
    Amplitude AA \rightarrow \infty. This correspondence to resonance.
  • Question 4
    1 / -0
    A simple wave motion represented by y=5(sin4πt+3cos4πt)y=5(\sin 4\pi t+\sqrt 3 \cos 4\pi t). Its amplitude is:
    Solution
    y=5(sin4πt+3cos4πt)y=5(sin4\pi t +\sqrt{3}cos4\pi t)
    =5×2(12sin4πt+32cos4πt)=5\times 2(\dfrac{1}{2}sin4\pi t+\dfrac{\sqrt{3}}{2}cos4\pi t)
    =10(cosπ3sin4πt+sinπ3cos4πt)=10(cos\dfrac{\pi}{3}sin4\pi t+sin\dfrac{\pi}{3}cos4\pi t)
    =10sin(4πt+π3)=10sin(4\pi t+\dfrac{\pi}{3})
    Hence the amplitude of the wave is 1010.
  • Question 5
    1 / -0
    The phase change between incident and reflected sound wave from a free end is
    Solution
    At a free and, the wave is reflected as it is with just as change in its direction of propagation 
    y=A sin (ωt+kx)y= A  \sin  ( \omega t+kx)
    yr=A sin (ωtkx)y_r= A  \sin  (\omega t-kx)
    phase difference in time domain =0= 0
  • Question 6
    1 / -0
    Which of the following is conserved when light waves interfere?
    Solution
    when light waves interfere energy is conserved. During interference energy is simply redistributed. Option C is correct.
  • Question 7
    1 / -0
    Thickness of very thin films can be found by the technique of
    Solution
    Ellipsometry is a technique that is used to measure properties of thin film.
    Ellipsometry uses the phenomenon of interference for the technique.
    Thus option B is correct answer.
  • Question 8
    1 / -0
    Phase difference between a particle at a compression and a particle at the next rarefaction is
    Solution
    Phase difference between two successive compression of rarefaction is 2π2\pi
    As  rarefaction appears between two compression, phase difference is π\pi .
  • Question 9
    1 / -0
    Two light waves are represented by y1=4sin ωty_{1}=4\sin  \omega t and y2=3sin(ωt+π2)y_{2}= 3\sin(\omega t+\frac{\pi }{2} ) . The resultant amplitude due to interference will be
    Solution
    The two waves are y1=4sinωty_{1}=4\sin \omega t and y2=3cosωty_{2}=3\cos \omega t, the resultant wave y=y1+y2y=y_{1}+y_{2}
    =4sinωt+3cosωt=4\sin \omega t + 3\cos \omega t.
    The resultant amplitude =42+32=\sqrt{4^{2}+3^{2}}
    =25=5=\sqrt{25}=5
  • Question 10
    1 / -0
    For the sustained interference of light, the necessary condition is that the two sources should
    Solution
    Interference refers to the interaction of waves that are coherent with each other , either because they come from the same source of because they have
    the same of nearly the same frequency. Thus option D is the correct option.
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