An expression of energy density is given by u = (α)/(β) (( )/(kt))...

Given that energy density $U = \dfrac{\alpha}{\beta}sin\dfrac{\alphax}{kt}$

As $\dfrac{\alphax}{kt}$ should be dimensionless, the dimensions of $\alphax and kt$ should be same.

$⇒ \left[\alpha\right]\left[x\right] = \left[k\right]\left[t\right] ⇒ \left[\alpha\right] = \dfrac{\left[k\right]\left[t\right]}{\left[x\right]}$

Now,

\left[U\right] = \dfrac{\left[\alpha\right]}{\left[\beta\right]}$</p><p>$⇒ \left[\beta\right] = \dfrac{\left[\alpha\right]}{\left[U\right]} = \dfrac{\left[k\right]\left[t\right]}{\left[x\right]\left[\dfrac{E}{V}\right]} = \dfrac{\left[\dfrac{E}{t}\right]\left[t\right]}{\left[x\right]\left[\dfrac{E}{V}\right]}$</p><p>$⇒ \left[\beta\right] = \dfrac{\left[{L}^{3}\right]}{\left[L\right]} = \left[{M}^{0}{L}^{2}{T}^{0}\right]