A string fastened at both ends resonates at 420 Hz and 490 Hz with no resonant frequencies in

Question

A string fastened at both ends resonates at $420 \mathrm{~Hz}$ and $490 \mathrm{~Hz}$ with no resonant frequencies in between. Find its fundamental resonant frequency. In general, $f_{n}=n f_{1}$. We are told that $f_{n}=420 \mathrm{~Hz}$ and $f_{n+1}=490 \mathrm{~Hz}$. Therefore, $$ 420 \mathrm{~Hz}=n f_{1} \text { and } 490 \mathrm{~Hz}=(n+1) f_{1} $$ Subtract the first equation from the second to obtain $f_{1}=70.0 \mathrm{~Hz}$.

          A string fastened at both ends resonates at $420 \mathrm{~Hz}$ and $490 \mathrm{~Hz}$ with no resonant frequencies in between. Find its fundamental resonant
frequency.
In general, $f_{n}=n f_{1}$. We are told that $f_{n}=420 \mathrm{~Hz}$ and $f_{n+1}=490 \mathrm{~Hz}$. Therefore,
$$
420 \mathrm{~Hz}=n f_{1} \text { and } 490 \mathrm{~Hz}=(n+1) f_{1}
$$
Subtract the first equation from the second to obtain $f_{1}=70.0 \mathrm{~Hz}$.

Added by Lynn J.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

04/11/2022

Step 1

Step 1: Given that $f_n = 420 \text{ Hz}$ and $f_{n+1} = 490 \text{ Hz}$, we have the equations $420 = n \cdot f_1$ and $490 = (n+1) \cdot f_1$. Show more…

Show all steps

Thanks for your feedback!

A string fastened at both ends resonates at $420 \mathrm{~Hz}$ and $490 \mathrm{~Hz}$ with no resonant frequencies in between. Find its fundamental resonant frequency. In general, $f_{n}=n f_{1}$. We are told that $f_{n}=420 \mathrm{~Hz}$ and $f_{n+1}=490 \mathrm{~Hz}$. Therefore, $$ 420 \mathrm{~Hz}=n f_{1} \text { and } 490 \mathrm{~Hz}=(n+1) f_{1} $$ Subtract the first equation from the second to obtain $f_{1}=70.0 \mathrm{~Hz}$.