Question

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

Answer 1

Hint: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 \[\left( {ABCD} \right)\]. 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 \[y = A\sin \left[ {Bx + Ct + D} \right]\].
Here unit of y is metre and its dimension is \[\left[ L \right]\].
We know that the trigonometric functions are dimensionless quantities so the term \[\sin \left[ {Bx + Ct + D} \right]\] 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 \[ = \left[ {{M^0}{L^1}{T^0}} \right]\]……(1)
The terms \[\left( {Bx} \right)\], \[\left( {Ct} \right)\] and D are adding quantities of a sine function so they are also dimensionless which can be expressed as \[\left[ {{M^0}{L^0}{T^0}} \right]\].
Dimension the term \[\left( {Bx} \right)\] is expressed as:
Dimension of \[\left( {Bx} \right) = \left[ {{M^0}{L^0}{T^0}} \right]\]……(2)
Here x is the displacement in x-direction, its unit is metre and dimension is \[\left[ L \right]\].
Substitute \[\left[ L \right]\] for the dimension of x in equation (2).
Dimension of \[\left( {Bx} \right) \cdot \left[ L \right] = \left[ {{M^0}{L^0}{T^0}} \right]\]
Dimension of B \[ = \left[ {{M^0}{L^{ - 1}}{T^0}} \right]\]
Dimension the term \[\left( {Ct} \right)\] is expressed as:
Dimension of \[\left( {Ct} \right) = \left[ {{M^0}{L^0}{T^0}} \right]\]……(3)
Here t is the time period, its unit is second and dimension is \[\left[ T \right]\].
Substitute \[\left[ T \right]\] for the dimension of t in equation (3).
Dimension of C\[ \cdot \left[ T \right] = \left[ {{M^0}{L^0}{T^{ - 1}}} \right]\]
Dimension of C \[ = \left[ {{M^0}{L^0}{T^{ - 1}}} \right]\]
Dimension of the term D is expressed as:
Dimension of D \[ = \left[ {{M^0}{L^0}{T^0}} \right]\]
Substitute \[\left[ {{M^0}{L^1}{T^0}} \right]\] for the dimension of A, \[\left[ {{M^0}{L^{ - 1}}{T^0}} \right]\] for the dimension of B, \[\left[ {{M^0}{L^0}{T^{ - 1}}} \right]\] for the dimension of C and \[\left[ {{M^0}{L^0}{T^0}} \right]\] for the dimension of D in dimensional formula for (ABCD).
Dimensional formula of (ABCD) \[\begin{array}{l}
 = \left[ {{M^0}{L^1}{T^0}} \right]\left[ {{M^0}{L^{ - 1}}{T^0}} \right]\left[ {{M^0}{L^0}{T^{ - 1}}} \right]\left[ {{M^0}{L^0}{T^0}} \right]\\
 = \left[ {{M^0}{L^0}{T^{ - 1}}} \right]
\end{array}\]
Therefore, the dimensional for (ABCD) is expressed as \[\left[ {{M^0}{L^0}{T^{ - 1}}} \right]\] 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.