On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is/are not correct.
| (a) | \(y = a\sin \left(2\pi t / T\right)\) |
| (b) | \(y = a\sin(vt)\) |
| (c) | \(y = \left({\dfrac a T}\right) \sin \left({\dfrac t a}\right)\) |
| (d) | \(y = a \sqrt 2 \left(\sin \left({\dfrac {2 \pi t} T}\right) - \cos \left({\dfrac {2 \pi t} T}\right)\right)\) |
(Symbols have their usual meanings.)
Choose the correct option:
| 1. | (a), (c) |
| 2. | (a), (b) |
| 3. | (b), (c) |
| 4. | (a), (d) |
Step: Analyse each option one by one.
Now, by using the principle of homogeneity of dimensions \(\text{LHS}\) and \(\text{RHS}\). of (\(\text{a}\)) and (\(\text{b}\)) will be the same, and is \(L\)
For the option (c), \(\text{[LHS]}=L\)
\(\begin{aligned} & [\text{RHS}]=\frac{L}{T}=[LT]^{-1} \\ & {[\text{LHS}] \neq[\text{RHS}]} \end{aligned}\)
Hence, (c) is not the correct option.
In option (b) the dimension of the angle is \(vt\) i.e., \(L\)
\(\begin{array}{cc} \Rightarrow & \text { RHS }=L \cdot L=L^2 \text { and } \text{LHS}=L \\ \Rightarrow & \text { LHS } \neq \text { RHS } \end{array}\)
So, option (b) is also not correct.
Therefore, options b and c are not correct.
Hence, option (3) is the correct answer.
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