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Waves Test - 39

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Waves Test - 39
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  • Question 1
    1 / -0
    During the propagation of wave motion, 
    Solution
    during propagation of waves energy  is transferred from one particle to to another without transfer of  particles. 
    so the answer is A.
  • Question 2
    1 / -0
    A particle in SHM is described by the displacement function: $$x(t)= A \ cos (\omega t+ \phi).$$ If the initial (t=0) position of the particle is 1 cm and its initial velocity is $$\pi $$ cm/sec , what is the amplitude?
    Solution
    $$x(t)=Acos(\omega t+\phi)$$
    $$\implies v(t)=\dfrac{dx(t)}{dt}$$
    $$=-\omega Asin(\omega t+\phi)$$
    Initial position=$$x(t=0)=Acos\phi=1cm$$
    Initial velocity=$$v(t=0)=-A\omega sin\phi=\omega cm/s$$
    Hence, $$A=\sqrt{2}cm$$
  • Question 3
    1 / -0
    The wave carries ______________
    Solution
    wave carries energy ,as propagation of wave is propagation of energy .
    so the answer is B.
  • Question 4
    1 / -0
    Which of the following is a Longitudinal wave? _____________
    Solution
    the example of longitudinal  waves is sound waves rest three are transverse in nature.
    so the answer is D.
  • Question 5
    1 / -0
    The displacement y of a wave traveling in the X-direction is given by $$y = 10^{-4}\sin({600t-2x+\dfrac{\pi}{3})}$$ metres. where $$x$$ is expressed in metres and $$t$$ in $$seconds$$. The speed of the wave motion in $$ms^{-1}$$ is:
    Solution
    $$y={ 10 }^{ -4 }sin\left( 600t-2x+\dfrac { \pi  }{ 3 }  \right) $$
    Here,  $$W=600/s$$
               $$K=2/m$$
    Speed of wave $$V=W/K$$
                                   $$=600/2$$
                                   $$=300m/s$$

    $$\therefore $$  Option (B) is correct.
  • Question 6
    1 / -0
    An excerpt from a book by Einstein and Infeld gives the following remarks concerning wave phenomena:
    "A bit of gossip starting in Washington reaches New York [by word of mouth] very quickly, even though not a single individual who takes part in spreading it travels between these two cities. There are two quite different motions involved, that of the rumor, Washington to New York, and that of the persons who spread the rumor."

    Identify a correct inference from the above text.
    Solution
    Sound waves are longitudinal waves, the particle itself do not travel from one place to other, they only vibrates in their positions and only energy is being transferred from one particle to other.
    Hence option (C) is correct.
  • Question 7
    1 / -0
    Three waves of amplitudes $$ 12\mu m $$, $$ 4\mu m $$ and $$ 9\mu m $$ but of same frequency arrive at a point in a medium with a successive phase diffeerence  of $$\dfrac{\pi}{2}$$. Then the resultant amplitude is
    Solution
    Since three waves have a successive phase difference of $$\pi/2$$.
    So, first and third waves have phase difference of $$\pi$$.
    Thus, net amplitude of first and third wave is  $$A_1 = 12-9 = 3\mu m$$
    Amplitude of second wave  $$A_2 = 4\mu m $$
    Resultant amplitude  $$A_R = \sqrt{A_1^2 + A_2^2}$$
    $$\therefore$$  $$A_R = \sqrt{3^2 + 4^2} = 5\mu m$$

  • Question 8
    1 / -0
    A traveling wave in a stretched string is described by the equation, $$y = A \sin(kx+t)$$. The maximum particle velocity is
    Solution
    $$v=\dfrac{dy}{dt}=Acos(kx+t)$$
    Since the max value of $$cos$$ is 1.
    The maximum velocity is $$A$$.
  • Question 9
    1 / -0
    Two sound waves each of amplitude $$a$$ and of frequencies $$500 Hz$$ and $$512 Hz$$ superpose. Then the resultant of the amplitude of the resultant wave at $$t=\dfrac{5}{48} s$$ is
    Solution
    When two sound waves of slightly different frequencies and equal amplitudes superimpose on each other, then their resultant amplitude is given by
    $$R= 2a\cos(2\pi\dfrac{n_1-n_2}{2}t)=2a\cos(2\pi\times\dfrac{512-500}{2}\times\dfrac{5}{48})=\dfrac{a}{\sqrt2}$$
    Note that here the angle is in radians.
    Substitute and calculate.
  • Question 10
    1 / -0
    If two waves of length $$50 cm$$ and $$51 cm$$ produced $$12$$ beats $$per\ second$$, the velocity of sound is:
    Solution
    Given,
    Wavelength,
    $${ \lambda  }_{ 1 }=50cm=0.5m\\ { \lambda  }_{ 2 }=51cm=0.51m$$
    Frequency of the beats, f=12
    Now, $$f=\frac { V }{ { \lambda  }_{ 1 } } -\frac { V }{ { \lambda  }_{ 2 } } =V\left[ \frac { 1 }{ { \lambda  }_{ 1 } } -\frac { 1 }{ { \lambda  }_{ 2 } }  \right] \\ 12=V\left[ \frac { 1 }{ 0.5 } -\frac { 1 }{ 0.51 }  \right] \\ 12=V\left[ \frac { 0.01 }{ 0.5\times 0.51 }  \right] \\ \left[ V=306m/s \right] $$
    Hence option (C) is correct
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