Dimensional Analysis
The displacement of a damped oscillator is described by y = A e^{-kt}, where t denotes time and A is an amplitude; what is the dimensional formula of the constant k?
Select the correct option:
Solution
[M0L0T−1]
A crucial rule governing transcendental functions is that the argument of an exponential, logarithm or trigonometric function must always be dimensionless, because such functions are defined through power series that add terms of different powers of the argument. In y = A e^(-kt), the exponent (-kt) must therefore be a pure number. Since t carries the dimension of time [T], the constant k must have dimensions that cancel it exactly, namely [T^-1], which is written fully as [M^0 L^0 T^-1]. The choice [T] is the inverse of the correct answer and would make the product kt have dimension [T^2], not dimensionless. The choice [L T^-1] is the dimension of velocity and has no place in a time-only exponent. The option labelled dimensionless confuses the requirement on the whole exponent with the dimension of k alone; the product kt is dimensionless, but k itself is not. This reasoning is the same applied to decay constants in NCERT chapters on oscillations and radioactivity. A check confirms it: k acts as the reciprocal of a characteristic decay time, naturally giving units of per second.
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About This Question
- Subject
- physics
- Chapter
- physics and measurement
- Topic
- dimensional analysis
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
[M0L0T−1]
A crucial rule governing transcendental functions is that the argument of an exponential, logarithm or trigonometric function must always be dimensionless, because such functions are defined through power series that add terms of different powers of the argument. In y = A e^(-kt), the exponent (-kt) must therefore be a pure number. Since t carries the dimension of time [T], the constant k must have dimensions that cancel it exactly, namely [T^-1], which is written fully as [M^0 L^0 T^-1]. The choice [T] is the inverse of the correct answer and would make the product kt have dimension [T^2], not dimensionless. The choice [L T^-1] is the dimension of velocity and has no place in a time-only exponent. The option labelled dimensionless confuses the requirement on the whole exponent with the dimension of k alone; the product kt is dimensionless, but k itself is not. This reasoning is the same applied to decay constants in NCERT chapters on oscillations and radioactivity. A check confirms it: k acts as the reciprocal of a characteristic decay time, naturally giving units of per second.
This medium difficulty physics question is from the chapter physics and measurement, covering the topic of dimensional analysis. It appeared in the 2025 exam.
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