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Waves Test - 74

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Waves Test - 74
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  • Question 1
    1 / -0
    A nylon guitar string has a linear density of 7.20 g/m7.20\ g/m and is under tension of 150 N150\ N. The fixed supports are distance D=90.0 cmD= 90.0\ cm apart. The string is oscillating in the standing wave pattern shown in figure.Calculate the
    (iii) The frequency of the traveling waves whose superposition gives this standing wave.

  • Question 2
    1 / -0
    A wave represented by y=100sin(ax+ bt)y = 100 \sin(ax +  bt) is reflected from a dense plane at the origin. If 3636% of energy is lost and rest of the energy is reflected then the equation of the reflected wave will be:-
    Solution

  • Question 3
    1 / -0
    Two waves of equal frequencies have their amplitudes in the ratio of 3 : 5. They are superimposed on each other. Calculate the ratio of maximum and minimum intensities of the resultant wave.
    Solution

  • Question 4
    1 / -0
    A particle executes SHM with a time period of 1616s. At time t=2st=2s, the particle crosses the mean position while at t=4st=4s, its velocity is 4ms14ms^{-1}. The amplitude of motion in metre is?
    Solution
    x=αsin(2π Tt+ϕ )x=\alpha sin\left( \dfrac { 2\pi  }{ T } t+\phi  \right)

    t=2s,x=0,T=16st=2s,x=0,T=16s

    0=αsin(π 4+ϕ )0=\alpha sin\left( \dfrac { \pi  }{ 4 } +\phi  \right)

    ϕ=π 4\phi =-\dfrac { \pi  }{ 4 }

    equ of shm is 

    x=asin(2π Ttπ 4 )x=asin\left( \dfrac { 2\pi  }{ T } t-\frac { \pi  }{ 4 }  \right)

    t=4s,v=4m/st=4s,v=4m/s

    v=dxdt=a×2π tcos(2π Ttπ 4 )v=\dfrac { dx }{ dt } =a\times \dfrac { 2\pi  }{ t } cos\left( \dfrac { 2\pi  }{ T } t-\dfrac { \pi  }{ 4 }  \right)

    4=α×2π 16cos(π 2π 4 )4=\alpha \times \dfrac { 2\pi  }{ 16 } cos\left( \dfrac { \pi  }{ 2 } -\dfrac { \pi  }{ 4 }  \right)

    4=α×2π 16×π 8×12 4=\alpha \times \dfrac { 2\pi  }{ 16 } \times \dfrac { \pi  }{ 8 } \times \dfrac { 1 }{ \sqrt { 2 }  }

    α=322 π \alpha =\dfrac { 32\sqrt { 2 }  }{ \pi  }
  • Question 5
    1 / -0
    If two waves, each of intensity I0{I}_{0}, having the same frequency but differing by a constant phase angle of 60o{60}^{o}, superpose at a certain point in space, then the intensity of resultant wave is:
    Solution

  • Question 6
    1 / -0
    A particle is executing SHM of amplitude AA, about the mean position x=0x = 0. Which of the following cannot be a possible phase difference between the positions of the particle at x=+A/2x = +A/2 and x=A/2x = -A/\sqrt {2}.
  • Question 7
    1 / -0
    Two particles execute S.H.M.S.H.M. along the same line at the same frequency. They move in opposite direction at the mean position. The phase difference will be:
    Solution

  • Question 8
    1 / -0
    Two waves are represented by x1=Asin(ωt+π6)x_1 = A \sin \left( \omega t + \dfrac{\pi}{6} \right) and x2=Acosωtx_2 = A \cos \omega t then the phase difference between them is :
    Solution

    The general equation of the waves given is

    x1=Asin(ωt+π6){x_1} = A\sin \left( {\omega t + \dfrac{\pi }{6}} \right)

    x2=Acos(ωt){x_2} = A\cos \left( {\omega t} \right)

    Then x1{x_1} can be written as

    x1=Acos(π2ωtπ6){x_1} = A\cos \left( {\dfrac{\pi }{2} - \omega t - \dfrac{\pi }{6}} \right)

    On comparing and solving we get the phase difference isπ3\dfrac{\pi }{3}

     

  • Question 9
    1 / -0
    Equation of a progressive wave is given by
    y=0.2cosπ0.04t+.02xπ6y = 0.2 \cos \pi \Bigg \lgroup 0.04t + .02x - \frac{\pi}{6} \Bigg \rgroup
    The distance is expressed in cm and time in second. What will be the minimum distance between two particles having the phase difference of π\pi/2
    Solution

  • Question 10
    1 / -0
    Two waves are propagating to the point p along a straight line produced by two sources A and B of ahead by π/3\pi /3 than that of B and the distance AP is greater than BP by 50 cm. Then the resultant amplitude at the point P will be, if the wavelength is 1 meter
    Solution

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