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In the equation y = A sin (omega t - kx)...

In the equation `y = A sin (omega t - kx),` obtain the dimensional formula of `omega and k.` Given x is distnace and t is time.

Text Solution

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The givenequation is `y =A sin (omega t - kx)`
The argument of a trigonometrical function, i.e.,
angle is dimensionless
`i.e., omega t =theta`
`omega = (theta)/(t) = (1)/(T) = T^(-1) = [M^0 L^0 T^(-1)]`
Also,`k x =theta`
`k =(theta)/(x) = (1)/(L) = L^(-1) = [M^0 L^(-1) T^0]`
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Knowledge Check

  • In the relation y = r sin ( omega t - kx) , the dimensions of omega//k are

    A
    `[M^(0) L^(0) T^(0)]`
    B
    `[M^(0) L^(1) T^(-1)]`
    C
    `[M^(0) L^(0) T^(1)]`
    D
    `[M^(0) L^(1) T^(0)]`

  • What are the dimensional formulae of omega and K in the relation y=Asin(omegat-Kx) ?

    A
    `[M^(0)L^(0)T^(2)][M^(0)L^(-1)T^(0)]`
    B
    `[M^(0)L^(0)T^(-1)][M^(0)L^(-1)T^(0)]`
    C
    `[M^(0)L^(0)T^(-1)][M^(0)L^(1)T^(0)]`
    D
    `[M^(0)L^(0)T^(-1)][M^(0)L^(1)T^(-1)]`

  • In the relation ( dy)/( dt) = 2 omega sin ( omega t + phi_(0)) , the dimensional formula for omega t + phi_(0) is

    A
    `MLT`
    B
    `MLT^(0)`
    C
    `ML^(0) T^(0)`
    D
    `M^(0) L^(0) T^(0)`

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