Dimensions Analysis

In a relation F \color{#D61F06}{F} = a sin ( k 1 x ) a\sin(k_1\color{#456461}{x}) + b sin ( k 2 T ) b\sin(k_2\color{#0C6AC7}{\text T}) where F , x F, x and T T denote the force , distance \color{#D61F06}{\text {force}} , \color{#456461}{\text {distance}} and Time \color{#0C6AC7}{\text{ Time}} respectively. Units of k 1 \color{magenta}{k_1} and k 2 \color{magenta}{k_2} respectively in S.I units are.

m , s 1 \text{m} , \text{s}^{-1} m 1 , s 1 \text{m}^{-1} , \text{s}^{-1} m 1 , s \text{m}^{-1} , \text{s} m , s \text{m} , \text{s}

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2 solutions

Kishore S. Shenoy
Aug 23, 2015

Angle's are dimensionless quantities.

k 1 x [ unit of k 1 ] m [ unit of k 1 ] = m 1 \Rightarrow k_1\cdot x \equiv [\text{unit of }k_1]\cdot m\\ \Rightarrow [\text{unit of }k_1] = m^{-1}

Similarly

k 2 T [ unit of k 2 ] s [ unit of k 1 ] = s 1 k_2\cdot T \equiv [\text{unit of }k_2]\cdot s\\ \Rightarrow [\text{unit of }k_1] = s^{-1}

Chew-Seong Cheong
Aug 21, 2015

sin ( k 1 x ) \sin{(k_1x)} and sin ( k 2 T ) \sin{(k_2T)} are scalars and without dimensions or units. Since the unit of x x is m m then that of k 1 k_1 is m 1 m^{-1} , and that of T T is s s and k 2 k_2 is s 1 s^{-1} . Therefore the answer is m 1 , s 1 \boxed{m^{-1}, s^{-1}} .

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