In the relation y - r sin -omega t - kx- the dimensions of omega-k are-M-0L-1T-0-M-0L-0T-1-M-0L-0T-0-M-0L-1-T-1-

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Toppr Admin · Accepted Answer

Y-rsin-wt-x2212-kx-wt-x2212-kr- is angle so it is dimensionless-wt-M0L0T0-kx-M0L0T0-w-1t-w-T-x2212-1-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- k-1xwk-T-x2212-1-L-x2212-1-M0LT-x2212-1-xA0- -xA0- -xA0- -xA0- -xA0-k-L-x2212-1-

In the relation $y = r s i n (ω t - k x)$ , the dimensions of $ω / k$ are
$[M^{0} L^{1} T^{0}]$
$[M^{0} L^{0} T^{1}]$
$[M^{0} L^{0} T^{0}]$
$[M^{0} L^{1} T^{- 1}]$

$y = r s i n (w t - k x)$
$(w t - k r)$ is angle so it is dimensionless.
$[w t] = [M^{0} L^{0} T^{0}] [k x] = [M^{0} L^{0} T^{0}]$
$w = \frac{1}{t}$
$[w] = [T^{- 1}]$ $k = \frac{1}{x}$
$\frac{w}{k} = \frac{[T^{- 1}]}{[L^{- 1}]} = [M^{0} {L T}^{- 1}]$ $k = [L^{- 1}]$

In the relation y=rsin(ωt−kx), the dimensions of ω/k are[M0L1T0][M0L0T1][M0L0T0][M0L1T−1]

y=rsin(wt−kx)(wt−kr) is angle so it is dimensionless.[wt]=[M0L0T0][kx]=[M0L0T0]w=1t[w]=[T−1] k=1xwk=[T−1][L−1]=[M0LT−1] k=[L−1]

In the relation $y = r s i n (ω t - k x)$ , the dimensions of $ω / k$ are
$[M^{0} L^{1} T^{0}]$
$[M^{0} L^{0} T^{1}]$
$[M^{0} L^{0} T^{0}]$
$[M^{0} L^{1} T^{- 1}]$

$y = r s i n (w t - k x)$
$(w t - k r)$ is angle so it is dimensionless.
$[w t] = [M^{0} L^{0} T^{0}] [k x] = [M^{0} L^{0} T^{0}]$
$w = \frac{1}{t}$
$[w] = [T^{- 1}]$ $k = \frac{1}{x}$
$\frac{w}{k} = \frac{[T^{- 1}]}{[L^{- 1}]} = [M^{0} {L T}^{- 1}]$ $k = [L^{- 1}]$