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Q2.14(d):
A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion : (y = (\frac{a}{\sqrt{2}}) (\sin \frac{2\pi t}{T} + \cos \frac{2\pi t}{T})) ((a) = maximum displacement of the particle, (v) = speed of the particle. (T) = time-period of motion). Rule out the wrong formulas on dimensional grounds.

Solution :

Dimensional Analysis of the Displacement Equation

To determine the validity of the equation $y = \frac{a}{\sqrt{2}} \left( \sin \frac{2\pi t}{T} + \cos \frac{2\pi t}{T} \right)$, we must apply the Principle of Homogeneity of Dimensions. This principle states that a physical equation is dimensionally correct if and only if the dimensions of all the terms on both sides of the equation are identical. [Per the Principle of Homogeneity].

Step 1: Identifying Dimensions of Given Variables

  • $y$: Displacement (Length) $\rightarrow [L]$
  • $a$: Maximum displacement (Amplitude/Length) $\rightarrow [L]$
  • $t$: Time $\rightarrow [T]$
  • $T$: Time-period $\rightarrow [T]$
  • $2\pi, \sqrt{2}$: Dimensionless constants.

Step 2: Analyzing the Trigonometric Arguments

The arguments of the trigonometric functions are $\frac{2\pi t}{T}$.

Dimensionally: $\frac{[T]}{[T]} = [M^0 L^0 T^0] = 1$.

Since the arguments are dimensionless, the functions $\sin(\theta)$ and $\cos(\theta)$ are valid and result in dimensionless quantities. [By the definition of Transcendental Functions in dimensional analysis].

Step 3: Checking Dimensional Consistency

The equation is: $y = \frac{a}{\sqrt{2}} \times (\text{dimensionless}) + \frac{a}{\sqrt{2}} \times (\text{dimensionless})$.

Dimensions of Right-Hand Side (RHS):

$[RHS] = [L] \times [M^0 L^0 T^0] = [L]$

Dimensions of Left-Hand Side (LHS):

$[LHS] = [L]$

Step 4: Vector-Scalar Diagrammatic Representation

Time (t) Displacement (y) y = (a/√2)(sin ωt + cos ωt)

Step 5: Theoretical Conclusion

Since $[LHS] = [L]$ and $[RHS] = [L]$, the dimensions match perfectly. The equation represents a simple harmonic motion with amplitude $A_{eff} = a$. Specifically, the expression can be rewritten as $y = a \sin\left(\frac{2\pi t}{T} + \frac{\pi}{4}\right)$, which confirms it is a physically valid expression for displacement.

Final Solution: The given formula $y = \frac{a}{\sqrt{2}} \left( \sin \frac{2\pi t}{T} + \cos \frac{2\pi t}{T} \right)$ is dimensionally correct.


More Questions from Class 11 Physics Units and Measurements EXERCISES


CBSE Solutions for Class 11 Physics Units and Measurements


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