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Waves Test - 70

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Waves Test - 70
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  • Question 1
    1 / -0
    A man $$x$$ can hear only upto $$10 kHz$$ and another man y upto $$20 Hz$$. A note of frequency $$500 Hz$$ is produced before them from a stretched string. Then
    Solution
    Both will hear sounds of same pitch and same quality
  • Question 2
    1 / -0
    In a plane progressive wave given by $$y= 25 \cos \left (2\pi t - \pi x \right )$$, the amplitude and frequency are respectively                              [BCECE 2003]

    Solution
    Comparing the given equation with $$y = a \cos \left ( \omega t - kx \right )$$
    $$a = 25, \omega = 2\pi t = 2\pi \Rightarrow n = 1 $$ Hz
  • Question 3
    1 / -0
    The phase difference between the two particles situated on both the side of a node is

    Solution
    We know that a node is formed when there is a stationary wave. A node is a point on the wave where the amplitude is always zero, which means that a nodal point never vibrates.
    When the two particles are in the two sides of a node their phase directions are completely opposite to each other. This means that the phase difference between them should be $$\pi$$ or $$180^{\circ}$$.
    Hence the correct answer is option (C). 
  • Question 4
    1 / -0
    Three waves of equal frequency having amplitudes $$10 \mu m, 4 \mu m$$ and $$7 \mu m$$  arrive at a given point with successive phase difference of $$\dfrac{\pi}{2}$$ The amplitude of the resulting wave in $$\mu m$$is given by 
    Solution
    The wave 1 and 3 reach out of phase. Hence resultant phase difference between them is $$\pi$$
    $$\therefore$$ 
    Resultant amplitude of 1 and 3 = 10 - 7 = 3 $$\mu m$$
    This wave has phase difference of$$\dfrac{\pi}{2}$$ with $$4 \mu m$$ 

     . 

  • Question 5
    1 / -0
    When a longitudinal wave propagates through a medium, the particles of the medium execute simple harmonic oscillations about their mean positions. These oscillations of a particle are characterised by an invariant           [SCRA 1998]
    Solution
    Total energy is conserved.
  • Question 6
    1 / -0
    Two waves having sinusoidal waveforms have different wavelengths and different amplitude. They will be having
    Solution
    The pitch depends upon the frequency of the source. As the two waves have different amplitude, therefore, they having different intensity. while quality depends on a number of harmonics/overtone produced and their relative intensity. Assuming that their frequencies are the same.
  • Question 7
    1 / -0
    A transverse progressive wave on a stretched string has a velocity of $$10ms^{-1}$$ and a frequency of 100 Hz. The phase difference between two particles of the string which are 23 cm apart will be
    Solution
    The phase difference of a wave is related to its path difference by the following relation,
    $$\triangle \phi = \dfrac{2\pi}{\lambda} \triangle x$$.....(i) 
    where $$\triangle\phi$$ is the phase difference, $$\triangle x$$ is the path difference and $$\lambda$$ is the wavelength. 
    In the question the velocity of the wave is given as $$v=10 m/s$$ and the frequency is given as $$\nu =100 Hz$$.
    The wavelength is given $$ v=\nu \lambda$$
    or, $$\lambda = \dfrac{v}{\nu}$$
    or, $$\lambda = 0.1m$$
    Using the value of wavelength in equation (i), we get,
    $$\triangle \phi=\dfrac{2\pi}{0.2} \times 2.3 \times \times 10^{-2}$$
    $$\triangle \phi = \dfrac{\pi}{2}$$ which is our required phase difference.
    The correct answer is option (D).
  • Question 8
    1 / -0
    Equation of motion in the same direction are given by 
    $$y_1 = 2a \sin \left (\omega t - kx  \right )$$ and $$y_1 = 2a \sin \left (\omega t - kx - \theta \right )$$
    The amplitude of the medium particle will be    [CPMT 2004]
    Solution
    When two waves are superpositioned, their resultant amplitude is given by,
    $$A_{new}=\sqrt{A_1^{2}+A_2^2+2A_1 A_2 cos \phi}$$, where $$\phi$$ is the phase difference.
    For these two waves we have $$A_1=A_2=a$$, and the phase difference is $$\theta$$
    $$\therefore a_{new}=\sqrt{a^2+a^2+2acos\theta}$$
    or, $$a_{new}=\sqrt{2a^2(1+cos\theta)}$$
    or, $$a_{new}=\sqrt{2} a\sqrt{1+cos\theta}$$
    We know that $$1+cos\theta = 2cos^2 \dfrac{\theta}{2}$$. So, using this,
    $$a_{new}=\sqrt{2} a \sqrt{(2cos^2\dfrac{\theta}{2})}$$
    or, $$A_{new}=2acos\dfrac{\theta}{2}$$, which is our required answer.
  • Question 9
    1 / -0
    In a wave, the path difference corresponding to a phase difference of $$\phi$$ is
    Solution
    For $$2 \pi$$ phase difference $$\rightarrow $$ Path difference is $$\lambda$$
    $$\therefore $$ For $$\phi$$ phase difference $$\rightarrow $$ Path difference is $$\dfrac{\lambda}{2 \pi} \phi$$
  • Question 10
    1 / -0
    Given in the graph above, the points $$A, B, C, D$$ represents state of vibration of a sound wave. From the below-mentioned options which represent the wavelength.

    Solution
    A) Distance between $$A$$ and $$C$$.
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