Standard +0.3 This is a standard dimensional analysis problem requiring students to equate dimensions of frequency with the given variables. While it involves multiple variables and solving simultaneous equations, the method is routine and commonly practiced in Further Mechanics. The algebra is straightforward with three unknowns and three equations from matching dimensions of M, L, and T.
A student conducts an experiment by first stretching a length of wire and fixing its ends. The student then plucks the wire causing it to vibrate. The frequency of these vibrations, \(f\), is modelled by the formula
$$f = kC^\alpha l^\beta \sigma^\gamma,$$
where \(C\) is the tension in the wire,
\(l\) is the length of the stretched wire,
\(\sigma\) is the mass per unit length of the stretched wire and
\(k\) is a dimensionless constant.
Use dimensional analysis to find \(\alpha\), \(\beta\) and \(\gamma\). [5]
Question 2:
2 | [f]=T −1 , [C]=MLT −2 and [σ]=ML −1 | B2 | 1.2
1.2 | B1 for any one correct
T −1 = ( MLT −2 )α L β( ML −1 )γ | M1 | 2.1 | [ ] [ ] [σ]
Using their f , C and to obtain
an equation in M, L and T
−2α=−1,α+β−γ=0,α+γ=0 | M1 | 1.1a | Setting up all three equations
1 1
α= ,β=−1,γ=−
2 2 | A1 | 1.1
[5]
A student conducts an experiment by first stretching a length of wire and fixing its ends. The student then plucks the wire causing it to vibrate. The frequency of these vibrations, $f$, is modelled by the formula
$$f = kC^\alpha l^\beta \sigma^\gamma,$$
where $C$ is the tension in the wire,
$l$ is the length of the stretched wire,
$\sigma$ is the mass per unit length of the stretched wire and
$k$ is a dimensionless constant.
Use dimensional analysis to find $\alpha$, $\beta$ and $\gamma$. [5]
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2020 Q2 [5]}}