Check the dimensional correctness of v = √(T/μ)
Speed of a wave on a string. T is the tension in the string and μ its mass per unit length.
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.
v = sqrt(T / μ)v[L T-1]=sqrt(T/mu)[L T-1]Step by step
1. What each symbol stands for
- v velocity [L T-1]
- T tension [M L T-2]
- μ linear mass density [M L-1]
2. The dimensions of each term
- v [L T-1] [L T-1]
- sqrt(T/mu) sqrt([M L T-2] / [M L-1]) [L T-1]
3. Compare
Every term has dimensions [L T-1], so the equation is dimensionally consistent.
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.
Matching dimensions are necessary but not sufficient. Pure numbers such as 1/2 or 2π have no dimensions, so a check cannot see them, and an equation with matching dimensions can still be wrong.
Look up each quantity
v = √(T/μ): quick answers
Is v = √(T/μ) dimensionally correct?
Yes. Every term has the same dimensions, [L T⁻¹], so v = √(T/μ) is dimensionally consistent. That does not prove the equation itself is correct.
Does a dimension check prove v = √(T/μ) 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.