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The question of a stationary wave is y=4...

The question of a stationary wave is `y=4cospix/5sin60pit` Where x and y are in cm and t in seconds. Find (a) the amplitude, (b) the velocity of component waves and © the distance between nodes

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Knowledge Check

  • The equation of a stationary wave is given by y=0.05cos(pix/3)sin40pit where y nad x are given in meter and time in sec, then the amplitude of each wave is,

    A
    5cm
    B
    2.5cm
    C
    10cm
    D
    3cm

  • The equation of stationary wave in stretched string is given by y=5sin(pix/3)cos40pit where x and y are in cm and t in sec. The separation between two adjacent node is,

    A
    1.5cm
    B
    3cm
    C
    6cm
    D
    4cm

  • The equation of stationary wave is y=3sinpix/20cospit (cm). The distance between antidote and next node is :

    A
    40cm
    B
    20cm
    C
    5cm
    D
    10cm

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    The equation of a stationary wave is given by y=0.03cos(2pix/0.03)sin15pit . The distance between on enode and one another node is given by

    The equation of a progressive wave is given by y=asin200pi(t-x/300) ,where x is in meter and t is in seconds. The frequency and velocity of the wave respectively

    The equation of a stationary wave is give by , y=0.05cos(2pix)sin(100pit). Find the frequency of waves. Here x and y are in S.I. units.

    The equation of sound wave is y=0.0015 sin.(64.4 x +316 t) Find the wavelength of the wave

    The displacement of a particle performing S.H.M. is given by x=3sin 5pit+ 4 cos5pit , where distance is in metre and time is in second. What is the amplitude of motion ?

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