Check the dimensional correctness of y = A sin(t/ω)
A sine with a time inside. The argument of a sine must be a pure number, and t/ω is a time squared.
Equation checker
Matching dimensions are necessary, not sufficient. A check cannot see pure numbers such as 1/2 or 2π, so a consistent equation can still be wrong.
y = A sin(t / ω)y[L]=A sin(t/omega)[L]Step by step
1. What each symbol stands for
- y displacement [L]
- A amplitude [L]
- t time [T]
- ω angular frequency [T-1]
2. The dimensions of each term
- y [L] [L]
- A sin(t/omega) [L] × sin([T] / [T-1]) [L]
3. Compare
sin( ) needs a plain number inside, an angle or a ratio, but "(t/omega)" has dimensions [T2].
Something inside a bracket does not match
A likely corrected form:
y = A sin(ωt - kx)
A wave moving along x. The angle inside the sine must be a pure number, and it is.
Check y = A sin(ωt - kx)What it proves and what it does not
It can show
- Every term that is added or subtracted has the same dimensions, so the sum makes sense.
- Both sides of the equals sign have the same dimensions.
- Angles inside sin, cos and similar functions are pure numbers.
It cannot show
- A pure number such as 1/2 or 2π has no dimensions, so a check cannot see a missing or wrong one.
- Two quantities with the same dimensions, such as work and torque, cannot be told apart.
- A sign error or a wrong power that keeps the dimensions equal goes unnoticed.
When the dimensions of the terms differ the equation cannot be correct, whatever numbers are in it, as long as each symbol means what the table says.
Look up each quantity
y = A sin(t/ω): quick answers
Is y = A sin(t/ω) dimensionally correct?
No. sin( ) needs a plain number inside, an angle or a ratio, but "(t/omega)" has dimensions [T²].
Does a dimension check prove y = A sin(t/ω) is right?
No. A check can show that an equation is wrong, but it cannot show that it is right. Pure numbers such as 1/2 and 2π have no dimensions, so the check cannot see them.