Answer to Question #279421 in Mechanics | Relativity for Ahmad

Question

A string with a length of 3 m has two adjacent resonances at frequencies 112 Hz and 140 Hz. Determine the speed of the waves on the string vibrating at 140 Hz.

Answer 1

Explanations & Calculations

• You need to use the formula of the harmonics of this arrangement.
• Let's assume the thread is fixed at both ends and only there.
• The formula would be then,

$\qquad\qquad \begin{aligned} \small f_n&=\small\frac{nv}{2L} \end{aligned}$

• Let's take the consecutive harmonics at $n\,\&\,(n+1)$ then,

$\qquad\qquad \begin{aligned} \small 112 &=\small \frac{nv}{2L}\implies n=\frac{112\times2L}{v}\cdots(1)\\ \\ \small 140&=\small\frac{(n+1)v}{2L}\implies n+1=\frac{140\times2L}{v}\cdots(2)\\ \\ \small n:(1)&\to(2),\\ \\ \small \frac{112\times2L}{v}+1&=\small \frac{140\times2L}{v}\\ \end{aligned}$

• Now you can substitute for the length of the string and give it a try to get the answer.

• A string in this arrangement can experience all the harmonics $\small n=1,2,3...$
• But we do not know what those given frequencies exactly corresponds to and that we used algebraic notation for that.
• The wave speed on a string remains constant as long as its tension, length and mass are kept unchanged.