Question:
A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is :
A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is :
Updated On: Jul 5, 2022
- 105 Hz
- 1.05 HZ
- 1050 HZ
- 10.5 HZ
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The Correct Option is A
Solution and Explanation
For string fixed at both the ends, resonant frequency are given by $f =\frac{n v}{2 L},$where symbols have their usual meanings. It is given that 315 Hz and 420 Hz are two consecutive resonant frequencies, let these are nth and (n + l)th harmonics.
$315=\frac{nv}{2L} \, \quad\quad\ldots\left(i\right)$
$420=\frac{\left(n+1\right)v}{2L} \, \quad\ldots$(ii)
$=E\left(i\right)+E\left(ii\right) \Rightarrow\frac{315}{420}=\frac{n}{n+1} \Rightarrow n=3$
From E (i), lowest resonant frequency
$f _{0}=\frac{v}{2L}=\frac{315}{3}=105 Hz$
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Waves
Waves are a disturbance through which the energy travels from one point to another. Most acquainted are surface waves that tour on the water, but sound, mild, and the movement of subatomic particles all exhibit wavelike properties. inside the most effective waves, the disturbance oscillates periodically (see periodic movement) with a set frequency and wavelength.
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Waves in which the medium moves at right angles to the direction of the wave.
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