## Related questions with answers

Two pieces of steel wire with identical cross sections have lengths of L and 2L. The wires are each fixed at both ends and stretched so that the tension in the longer wire is four times greater than in the shorter wire. If the fundamental frequency in the shorter wire is 60 Hz, what is the frequency of the second harmonic in the longer wire?

Solutions

VerifiedIn this problem, two steel wires of the same cross sectional area have lengths $L_\text{short} = L$ and $L_\text{long} = 2L$. The ends of the wires are fixed, and $F_\text{long} = 4F_\text{short}$. The fundamental frequency of the shorter wire is $f_\text{1, short} = 60~\mathrm{Hz}$. We calculate the second harmonic of the longer wire, $f_\text{2,long}$.

Lets look at the expression for the frequency of the second harmonic of the string with length $2L$ and tension $4F$ :

$\begin{align*} f'_2&=\frac{2}{2(2L)}\sqrt{\frac{4F}{\mu}}\\ &=\frac{1}{2L}\left(2\sqrt{\frac{F}{\mu}}\right)\\ &=2\frac{1}{2L}\sqrt{\frac{F}{\mu}}\\ &=2f_1\\ &=2(60\,\text{Hz})\\ &=\boxed{120\,\text{Hz}} \end{align*}$

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th Edition•ISBN: 9780133942651 (8 more)Randall D. Knight#### Mathematical Methods in the Physical Sciences

3rd Edition•ISBN: 9780471198260 (1 more)Mary L. Boas#### Fundamentals of Physics

10th Edition•ISBN: 9781118230718 (3 more)David Halliday, Jearl Walker, Robert Resnick## More related questions

1/4

1/7