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Question: Given that the displacement of an oscillating particle is given by \[y = A\sin \left[ {Bx + Ct + D} ...

Given that the displacement of an oscillating particle is given by y=Asin⁡[Bx+Ct+D]y = A\sin \left[ {Bx + Ct + D} \right]. The dimensional formula for (ABCD)\left( {ABCD} \right) is
A. M0L−1T0{M^0}{L^{ - 1}}{T^0}
B. M0L0T−1{M^0}{L^0}{T^{ - 1}}
C. M0L−1T−1{M^0}{L^{ - 1}}{T^{ - 1}}
D. M0L0T0{M^0}{L^0}{T^0}

Explanation

Solution

In the solution of this problem we will be calculating the individual dimension of A, B, C and D to evaluate the dimensional formula for (ABCD)\left( {ABCD} \right). It is also known to us that trigonometric functions such as sin, cos, tan, etc. are dimensionless.

Complete step by step answer:
It is given that the displacement of an oscillating particle is y=Asin⁡[Bx+Ct+D]y = A\sin \left[ {Bx + Ct + D} \right].
Here unit of y is metre and its dimension is [L]\left[ L \right].
We know that the trigonometric functions are dimensionless quantities so the term sin⁡[Bx+Ct+D]\sin \left[ {Bx + Ct + D} \right] in displacement y of an oscillating particle is a dimensionless quantity.
The dimension of A is equal to the dimension of y.
Dimension of A =[M0L1T0] = \left[ {{M^0}{L^1}{T^0}} \right]……(1)
The terms (Bx)\left( {Bx} \right), (Ct)\left( {Ct} \right) and D are adding quantities of a sine function so they are also dimensionless which can be expressed as [M0L0T0]\left[ {{M^0}{L^0}{T^0}} \right].
Dimension the term (Bx)\left( {Bx} \right) is expressed as:
Dimension of (Bx)=[M0L0T0]\left( {Bx} \right) = \left[ {{M^0}{L^0}{T^0}} \right]……(2)
Here x is the displacement in x-direction, its unit is metre and dimension is [L]\left[ L \right].
Substitute [L]\left[ L \right] for the dimension of x in equation (2).
Dimension of (Bx)⋅[L]=[M0L0T0]\left( {Bx} \right) \cdot \left[ L \right] = \left[ {{M^0}{L^0}{T^0}} \right]
Dimension of B =[M0L−1T0] = \left[ {{M^0}{L^{ - 1}}{T^0}} \right]
Dimension the term (Ct)\left( {Ct} \right) is expressed as:
Dimension of (Ct)=[M0L0T0]\left( {Ct} \right) = \left[ {{M^0}{L^0}{T^0}} \right]……(3)
Here t is the time period, its unit is second and dimension is [T]\left[ T \right].
Substitute [T]\left[ T \right] for the dimension of t in equation (3).
Dimension of C⋅[T]=[M0L0T−1] \cdot \left[ T \right] = \left[ {{M^0}{L^0}{T^{ - 1}}} \right]
Dimension of C =[M0L0T−1] = \left[ {{M^0}{L^0}{T^{ - 1}}} \right]
Dimension of the term D is expressed as:
Dimension of D =[M0L0T0] = \left[ {{M^0}{L^0}{T^0}} \right]
Substitute [M0L1T0]\left[ {{M^0}{L^1}{T^0}} \right] for the dimension of A, [M0L−1T0]\left[ {{M^0}{L^{ - 1}}{T^0}} \right] for the dimension of B, [M0L0T−1]\left[ {{M^0}{L^0}{T^{ - 1}}} \right] for the dimension of C and [M0L0T0]\left[ {{M^0}{L^0}{T^0}} \right] for the dimension of D in dimensional formula for (ABCD).
Dimensional formula of (ABCD) =[M0L1T0][M0L−1T0][M0L0T−1][M0L0T0] =[M0L0T−1]\begin{array}{l} = \left[ {{M^0}{L^1}{T^0}} \right]\left[ {{M^0}{L^{ - 1}}{T^0}} \right]\left[ {{M^0}{L^0}{T^{ - 1}}} \right]\left[ {{M^0}{L^0}{T^0}} \right]\\\ = \left[ {{M^0}{L^0}{T^{ - 1}}} \right] \end{array}
Therefore, the dimensional for (ABCD) is expressed as [M0L0T−1]\left[ {{M^0}{L^0}{T^{ - 1}}} \right] and option (B) is correct.

Note: Try to remember the concept of homogeneity of dimensions while performing basic arithmetic operations such as addition, subtraction, multiplication and division.