The force F is given in terms of time t and d | vedclass

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Question

$F=A \cos B x+C \sin D t$

the argument, $	heta$ of cos or sin should be dimensionless.

therefore,

dimension of $\mathrm{Bx}=[M L T]$

$[B]

Vedclass · Accepted Answer

${M^0}{L^1}{T^{ - 1}}$

Vedclass · Answer

$F=A \cos B x+C \sin D t$

the argument, $	heta$ of cos or sin should be dimensionless.

therefore,

dimension of $\mathrm{Bx}=[M L T]$

$[B]\left[L^{\prime}ight]=[M L T]$

$[B]=\left[M L^{0} Tight]$

Similarly $[D]\left[T^{\prime}ight]=[M L T]$

$[D]=\left[M L T^{0}ight.$

dimension of $D/B=\frac{\left[M L T^{0}ight]}{\left[M L^{0} Tight]}$

$=\left[L^{1} T^{-1}ight]$

The force $F$ is given in terms of time $t$ and displacement $x$ by the equation $F = A\,cos\,Bx + C\,sin\,Dt.$ The dimensional formulae of $D/B$ is

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