The force F is given by expression F=A cos (B x)+C sin (D t), where x is the displacement and t is the time. Then dimension of (D/B) are same as that of

Question

The force F is given by expression F=A cos (B x)+C sin (D t), where x is the displacement and t is the time. Then dimension of (D/B) are same as that of (A) velocity [LT-1] (B) angular velocity [T-1] (C) angular momentum [ML2T-1] (D) velocity gradient [T-1]

Tardigrade Admin · Accepted Answer

In trigonometric function like sin θ, cos θ etc θ is dimensionless So, .[B x]= M 0 L0 T0] [B]=[ L -1] Also, [Dt] = [M0L0T0] [D] = [T-1] Now [(D/B)]=[ LT -1]

Q. The force $F$ is given by expression $F = A cos (B x) + C sin (D t)$ , where $x$ is the displacement and $t$ is the time. Then dimension of $\frac{D}{B}$ are same as that of

velocity $[L T^{- 1}]$

angular velocity $[T^{- 1}]$

angular momentum $[M L^{2} T^{- 1}]$

velocity gradient $[T^{- 1}]$

In trigonometric function like
$sin θ, cos θ$ etc
$θ$ is dimensionless
So, $[B x] = M^{0} L^{0} T^{0}]$
$[B] = [L^{- 1}]$
Also, $[D t] = [M^{0} L^{0} T^{0}]$
$[D] = [T^{- 1}]$
Now $[\frac{D}{B}] = [L T^{- 1}]$

Q. The force F is given by expression F=Acos(Bx)+Csin(Dt), where x is the displacement and t is the time. Then dimension of BD​ are same as that of

velocity [LT−1]

angular velocity [T−1]

angular momentum [ML2T−1]

velocity gradient [T−1]