Question

A string vibrates in five segments at a frequency of $460 \mathrm{~Hz} .(a)$ What is its fundamental frequency? (b) What frequency will cause it to vibrate in three segments? If the string is $n$ segments long, then fromwe have $n\left(\frac{1}{2} \lambda\right)$ $=L$. But $\lambda=v / f_{n}$, so $L=n\left(\mathrm{u} / 2 f_{n}\right)$. Solving for $f_{n}$ provides $$ f_{n}=n\left(\frac{v}{2 L}\right) $$ We are told that $f_{5}=460 \mathrm{~Hz}$, and so $$ 460 \mathrm{~Hz}=5\left(\frac{v}{2 L}\right) \quad \text { or } \quad \frac{v}{2 L}=92.0 \mathrm{~Hz} $$ Substituting this in the above relation gives $$ f_{n}=(n)(92.0 \mathrm{~Hz}) $$ (a) $f_{1}=92.0 \mathrm{~Hz}$. (b) $f_{3}=(3)(92 \mathrm{~Hz})=276 \mathrm{~Hz}$ Alternative Method Recall that for a string held at both ends, $f_{n}=n_{f 1}$. Knowing that $f_{5}$ $=460 \mathrm{~Hz}$, it follows that $f_{1}=92.0 \mathrm{~Hz}$ and $f_{3}=276 \mathrm{~Hz}$.

          A string vibrates in five segments at a frequency of $460 \mathrm{~Hz} .(a)$ What is its fundamental frequency? (b) What frequency will cause it to vibrate in three segments?
If the string is $n$ segments long, then fromwe have $n\left(\frac{1}{2} \lambda\right)$ $=L$. But $\lambda=v / f_{n}$, so $L=n\left(\mathrm{u} / 2 f_{n}\right)$. Solving for $f_{n}$ provides
$$
f_{n}=n\left(\frac{v}{2 L}\right)
$$
We are told that $f_{5}=460 \mathrm{~Hz}$, and so
$$
460 \mathrm{~Hz}=5\left(\frac{v}{2 L}\right) \quad \text { or } \quad \frac{v}{2 L}=92.0 \mathrm{~Hz}
$$
Substituting this in the above relation gives
$$
f_{n}=(n)(92.0 \mathrm{~Hz})
$$
(a) $f_{1}=92.0 \mathrm{~Hz}$.
(b) $f_{3}=(3)(92 \mathrm{~Hz})=276 \mathrm{~Hz}$
Alternative Method
Recall that for a string held at both ends, $f_{n}=n_{f 1}$. Knowing that $f_{5}$ $=460 \mathrm{~Hz}$, it follows that $f_{1}=92.0 \mathrm{~Hz}$ and $f_{3}=276 \mathrm{~Hz}$.

Added by Patty M.

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