Let a wave be described by the equation y=asinω ((x/v) - k) where, ω is angular velocity and v is the linear velocity. What will be the dimensions of k?

Question

Let a wave be described by the equation y=asinω ((x/v) - k) where, ω is angular velocity and v is the linear velocity. What will be the dimensions of k? (A) [T2] (B) [T- 1] (C) [T] (D) [L T]

Tardigrade Admin · Accepted Answer

y=asin((ω x/v) - ω k) Angle is dimensionless, so [ω k]=M0L0T0 i.e. [k]=(1/[ω ])=(1/T- 1)=T

Q. Let a wave be described by the equation
$y = a s inω (\frac{x}{v} - k)$
where, $ω$ is angular velocity and $v$ is the linear velocity. What will be the dimensions of $k ?$

$[T^{2}]$

$[T^{- 1}]$

$[T]$

$[L T]$

$y = a s in (\frac{ω x}{v} - ωk)$
Angle is dimensionless, so
$[ωk] = M^{0} L^{0} T^{0}$
i.e. $[k] = \frac{1}{[ ω ]} = \frac{1}{T ^{- 1}} = T$

Q. Let a wave be described by the equation y=asinω(vx​−k) where, ω is angular velocity and v is the linear velocity. What will be the dimensions of k?

[T2]

[T−1]

[T]

[LT]

y=asin(vωx​−ωk) Angle is dimensionless, so [ωk]=M0L0T0 i.e. [k]=[ω]1​=T−11​=T

Q. Let a wave be described by the equation
$y = a s inω (\frac{x}{v} - k)$
where, $ω$ is angular velocity and $v$ is the linear velocity. What will be the dimensions of $k ?$

$[T^{2}]$

$[T^{- 1}]$

$[T]$

$[L T]$

$y = a s in (\frac{ω x}{v} - ωk)$
Angle is dimensionless, so
$[ωk] = M^{0} L^{0} T^{0}$
i.e. $[k] = \frac{1}{[ ω ]} = \frac{1}{T ^{- 1}} = T$