Determining Dimensional Correctness of Physics Equations: Velocity and Trigonometric Analysis
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Reference
Serway & Jewett's "Physics for Scientists and Engineers with Modern Physics 9th Edition" is a comprehensive textbook that expertly covers classical mechanics, electromagnetism, thermodynamics, and modern physics concepts, providing students with detailed explanations, practical examples, and rigorous problem-solving approaches in physics.
Question
9. Which of the following equations are dimensionally correct? (a) vf = vi + ax (b) y = (2 m) cos (kx), where k = 2 m⁻¹
Explain Question
This question tests understanding of dimensional analysis in physics equations, requiring us to examine two different equations for dimensional consistency. For equation (a), we need to analyze the dimensions of velocity, acceleration, and distance terms, while for equation (b), we must consider the dimensions of length and trigonometric functions with their arguments.
Explain Solve
For equation (a), we first identify the dimensions of each term: final velocity (vf) and initial velocity (vi) have dimensions of LT⁻¹, acceleration (a) has dimensions of LT⁻², and distance (x) has dimensions of L. We then compare the dimensions on both sides of the equation to check for consistency.
For equation (b), we analyze the dimensions of length (y and m), the dimensionless nature of cosine functions, and ensure the argument of cosine (kx) is dimensionless. The key insight is that trigonometric functions require dimensionless arguments to be mathematically valid.
Summary
This problem demonstrates the importance of dimensional analysis in physics, showing how equations must maintain dimensional consistency to be physically meaningful. The analysis reveals that proper dimensionality is crucial in both mechanical equations and trigonometric expressions involving physical quantities.
Solve
For equation (a): The left side has dimensions [vf] = LT⁻¹, while the right side combines [vi] = LT⁻¹ and [ax] = L²T⁻², making the equation dimensionally incorrect as we cannot add terms with different dimensions.
For equation (b): The left side [y] = L matches the right side [(2 m)cos(kx)] = L × 1, where kx is dimensionless (L⁻¹ × L = 1), making this equation dimensionally correct as the cosine of a dimensionless quantity multiplied by a length gives a length.
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