Q.(MCQ II — one or more options correct) On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct:

The key idea is Dimensional Analysis.

For a physical relation to be correct, it must satisfy the principle of dimensional homogeneity:

1. The dimensions of all terms on both sides of the equation must be identical.
2. Arguments of trigonometric functions (like $\sin$, $\cos$, $\tan$) must be dimensionless.

Let's check each option:

• Displacement $y$ and amplitude $a$ have dimension $[L]$.
• Time $t$ and time period $T$ have dimension $[T]$.
• Velocity $v$ has dimension $[LT^{-1}]$.

(A) $y = a \sin \dfrac{2\pi t}{T}$:

* LHS: $[y] = [L]$

* RHS: $[a] = [L]$. The argument $\left[\dfrac{2\pi t}{T}\right] = \dfrac{[T]}{[T]} = [M^0 L^0 T^0]$ (dimensionless).

* This relation is dimensionally correct.

(B) $y = a \sin vt$:

* LHS: $[y] = [L]$

* RHS: $[a] = [L]$. The argument $[vt] = [LT^{-1}][T] = [L]$. This is not dimensionless.

* This relation is dimensionally incorrect.

(C) $y = \dfrac{a}{T} \sin\left(\dfrac{t}{a}\right)$:

* LHS: $[y] = [L]$

* RHS: $\left[\dfrac{a}{T}\right] = \dfrac{[L]}{[T]} = [LT^{-1}]$. The dimensions on LHS and RHS do not match. …

When we analyze physical equations, one of the most fundamental checks we can perform is dimensional analysis. This method ensures that the equation is consistent with the units of the physical quantities involved. If an equation is dimensionally incorrect, it cannot possibly describe a physical phenomenon accurately, regardless of the numerical values.

The core idea behind dimensional analysis rests on two principles:

1. Homogeneity Principle: The dimensions of all terms on both sides of an equation must be identical. You cannot add or subtract quantities with different dimensions (e.g., you can't add a length to a time).
2. Dimensionless Arguments: The arguments of transcendental functions (like trigonometric functions such as $\sin$, $\cos$, $\tan$; exponential functions $e^x$; and logarithmic functions $\ln x$) must always be dimensionless. This is because these functions are defined by series expansions (e.g., $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$), and for such an expansion to be dimensionally consistent, $x$ must be dimensionless. If $x$ had dimensions, then $x$ and $x^3$ would have different dimensions, making the sum impossible.

For this problem, we are dealing with displacement ($y$) in simple harmonic motion (SHM). Let's establish the dimensions of the quantities involved:

• Displacement, $y$: $[L]$ (Length)
• Amplitude, $a$: $[L]$ (Length)
• Time, $t$: $[T]$ (Time)
• Period, $T$: $[T]$ (Time)
• Velocity, $v$: $[LT^{-1}]$ (Length per Time)
• $2\pi$ and $\sqrt{2}$ are pure numbers, hence dimensionless.

Now, let's examine each option:

1. Option (A): $y = a \sin \dfrac{2\pi t}{T}$

   • Left side: The dimension of $y$ is $[L]$.
   • Right side:
     • The dimension of $a$ is $[L]$.
     • Consider the argument of the sine function: $\dfrac{2\pi t}{T}$.
       • Dimensions of $2\pi$: Dimensionless.
       • Dimensions of $t$: $[T]$.
       • Dimensions of $T$ (period): $[T]$.
       • So, the dimension of the argument is $\dfrac{[T]}{[T]} = [T^0]$, which is dimensionless. This is correct.
     • Since the argument is dimensionless, $\sin\left(\dfrac{2\pi t}{T}\right)$ is also dimensionless.
     • Therefore, the dimension of the right side is $[L] \times [T^0] = [L]$.
   • Conclusion: The dimensions on both sides match ($[L] = [L]$). This relation is dimensionally correct.

2. Option (B): $y = a \sin vt$

   • Left side: The dimension of $y$ is $[L]$.
   • Right side:
     • The dimension of $a$ is $[L]$.
     • Consider the argument of the sine function: $vt$.
       • Dimensions of $v$: $[LT^{-1}]$.
       • Dimensions of $t$: $[T]$.
       • So, the dimension of the argument is $[LT^{-1}] \times [T] = [L]$.
     • Problem: The argument $[L]$ is not dimensionless. This violates the rule that arguments of trigonometric functions must be dimensionless.
   • Conclusion: This relation is dimensionally incorrect.
   Watch out

   A common mistake is to only check the overall dimensions of the equation. Always remember to check the arguments of trigonometric, exponential, and logarithmic functions; they must be dimensionless.

3. Option (C): $y = \dfrac{a}{T} \sin\left(\dfrac{t}{a}\right)$

   • Left side: The dimension of $y$ is $[L]$.
   • Right side:
     • Consider the term outside the sine function: $\dfrac{a}{T}$.
       • Dimensions of $a$: $[L]$.
       • Dimensions of $T$: $[T]$.
       • So, the dimension of $\dfrac{a}{T}$ is $\dfrac{[L]}{[T]} = [LT^{-1}]$.
     • Consider the argument of the sine function: $\dfrac{t}{a}$.
       • Dimensions of $t$: $[T]$. …

Method: Dimensional Analysis

Concept: For any valid physical equation, the dimensions on both sides must match. In SHM, displacement $y$ has dimensions of length $[L]$, and the argument of sine (or cosine) must be dimensionless.

Steps

1. Write dimensions of known quantities

   • $y$ (displacement) → $[L]$
   • $a$ (amplitude) → $[L]$
   • $t$ (time) → $[T]$
   • $T$ (time period) → $[T]$
   • $v$ (velocity) → $[LT^{-1}]$

2. Check each option

   Option (A): $y = a \sin \dfrac{2\pi t}{T}$

   • Argument: $\dfrac{2\pi t}{T}$ → $\dfrac{[T]}{[T]} = \text{dimensionless}$ ✓
   • RHS: $[L] \times \text{(dimensionless)} = [L]$ ✓
   • Correct relation

   Option (B): $y = a \sin vt$

   • Argument: $vt$ → $[LT^{-1}] \times [T] = [L]$ ✗
   • Sine of a length is meaningless.
   • Not correct

   Option (C): $y = \dfrac{a}{T} \sin\left(\dfrac{t}{a}\right)$ …

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting that the argument of sine/cosine must be dimensionless

The error: Students check only the amplitude side ($y$ vs $a$) and ignore the sine argument. In option (C), $y = \dfrac{a}{T} \sin\left(\dfrac{t}{a}\right)$, the argument $\dfrac{t}{a}$ has dimensions of $\dfrac{[T]}{[L]}$ — not dimensionless. This alone makes it invalid.

How to avoid: Always check the argument of every trigonometric, exponential, or logarithmic function first. It must be a pure number (dimensionless). Write:

• $[\text{argument}] = [M^0 L^0 T^0]$

Mistake 2: Assuming $v$ in $\sin(vt)$ is velocity

The error: In option (B), $y = a \sin(vt)$, students think $v$ is velocity. But $[vt] = [LT^{-1}][T] = [L]$, which has dimensions of length — not dimensionless. So this is dimensionally wrong.

How to avoid: Never assume a symbol means what it "usually" means in physics. In dimensional analysis, treat every symbol by its given dimensions:

• $v$ here has dimensions of velocity $[LT^{-1}]$
• So $vt$ has dimensions of length, not dimensionless

Mistake 3: Thinking "amplitude must have same dimensions as displacement" is enough

The error: In option (C), $y = \dfrac{a}{T} \sin(\dots)$, the amplitude $\dfrac{a}{T}$ has dimensions $[LT^{-1}]$, but displacement $y$ must have $[L]$. So even if the argument were correct, the amplitude is wrong.

How to avoid: Check both:

1. Amplitude dimensions must match $y$ (displacement)
2. Argument of sine/cosine must be dimensionless

Mistake 4: Missing that option (D) is actually correct

The error: Students see $\sqrt{2}$ and think it's "suspicious" or dismiss it. But $\sqrt{2}$ is dimensionless, and the argument $\dfrac{2\pi t}{T}$ is dimensionless. The amplitude $a\sqrt{2}$ has dimensions $[L]$, matching $y$.

How to avoid: Don't reject a relation just because it looks "unusual." Check dimensions systematically:

• $[a\sqrt{2}] = [L]$ ✓ …