MATH SOLVE

2 months ago

Q:
# the pitch, or frequency, of a vibrating string varies directly with the square root of the tension. if a string vibrates at a frequency of 300 hertz due to a tension of 8 pounds, find the frequency when the tension is 72 pounds.

Accepted Solution

A:

[tex]\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\
\begin{array}{llll}
\stackrel{frequency}{f}\textit{ of a vibrating string varies directly}\\
\qquad \qquad \textit{with the square root of the tension }\stackrel{tension}{t}
\end{array}[/tex]

[tex]\bf f=k\sqrt{t}\quad \textit{we also know that } \begin{cases} f=300\\ t=8 \end{cases}\implies 300=k\sqrt{8} \\\\\\ \cfrac{300}{\sqrt{8}}=k\implies \cfrac{300}{\sqrt{2^2\cdot 2}}=k\implies \cfrac{300}{2\sqrt{2}}=k\implies \cfrac{150}{\sqrt{2}}=k \\\\\\ \textit{and we can \underline{rationalize} it to }\cfrac{150\sqrt{2}}{2}\implies 75\sqrt{2}=k[/tex]

[tex]\bf thus\qquad f=\stackrel{k}{75\sqrt{2}}\sqrt{t}\implies \boxed{f=75\sqrt{2t}}\\\\ -------------------------------\\\\ \textit{now, when t = 72, what is \underline{f}?}\qquad f=75\sqrt{2(72)}[/tex]

[tex]\bf f=k\sqrt{t}\quad \textit{we also know that } \begin{cases} f=300\\ t=8 \end{cases}\implies 300=k\sqrt{8} \\\\\\ \cfrac{300}{\sqrt{8}}=k\implies \cfrac{300}{\sqrt{2^2\cdot 2}}=k\implies \cfrac{300}{2\sqrt{2}}=k\implies \cfrac{150}{\sqrt{2}}=k \\\\\\ \textit{and we can \underline{rationalize} it to }\cfrac{150\sqrt{2}}{2}\implies 75\sqrt{2}=k[/tex]

[tex]\bf thus\qquad f=\stackrel{k}{75\sqrt{2}}\sqrt{t}\implies \boxed{f=75\sqrt{2t}}\\\\ -------------------------------\\\\ \textit{now, when t = 72, what is \underline{f}?}\qquad f=75\sqrt{2(72)}[/tex]