Waves

Let the velocity of the wave in the string be given by $v$ , and let the length of the string be given by $L$ . The wave in a string is formed by both it's ends as nodes. It can be easily seen that, the wavelength, $\lambda$ , of the wave formed is given by ${\lambda}_{n} = \frac {2L} {n}$ , where $n$ is a natural number.

Now, according to the given question, since there is no other frequency between $420 Hz$ and $315 Hz$ , hence these frequencies can be written as

${\nu}_{n} = \frac {v} { {\lambda}_{n} } =420 Hz$

and ${\nu}_{n-1} = \frac {v} { {\lambda}_{n-1} } =315 Hz$

So, $\frac {v n} {2L} =420$ and $\frac {v (n-1)} {2L} = 315$

Solving, the equations, we get $n=4$

Now, the fundamental frequency for a string is given by $\frac {v} {2L}$ which can be calculated by using the value of $n$ .

So, solving the formed equations: $\frac {v} {2L} = 105 Hz$

This explanation is written in a way which will (hopefully) be useful for those studying A-Level Physics.

From earlier study of harmonics on stationary waves, we know that the first harmonic frequency is $f_{1}$ = $\frac{c}{λ_{1}}$ , where $c$ is the wavespeed on the string. We do not need the value of $c$ to solve this question, but for reference, it can be calculated as $\frac{T}{μ} ^{0.5}$ , where $T$ is the tension in the string, and $μ$ is the mass per unit length in the string. The frequency of the first harmonic $f_{1}$ can be calculated as $f_{1} = \frac{c}{2L}$ , where $L$ is the length of the string, because $2L = λ_{1}$ . You can visualise this by looking at a stationary wave at the first harmonic: there is one "loop" (equal to half a wavelength) between the two fixed ends, meaning that the wavelength $λ_{1}$ is twice the distance between the ends of the string.

For the second harmonic frequency, $f_{2}$ , there are two "loops" between the ends of the string, or one wavelength. Therefore, $f_{2}$ = $\frac{c}{L}$ , as $f_{2}$ = $\frac{c}{λ_{2}}$ . Rearranging shows that $f_{2} = 2f_{1}$ . It turns out that for the nth harmonic stationary wave, the frequency $f_{n} = nf_{1}$ .

We will take the frequency $f_{n} = 315$ Hz. It is given that there are no harmonics (resonant freqencies) between the two frequencies in the question, so it follows that $f_{n+1} = 420$ Hz. Therefore: $nf_{1} = 315$ $(n+1)f_{1} = 420$ ∴ $nf_{1} + f_{1} = 420$ ∴ $f_{1} = 420 - 315 = 105$ ∴ $f_{1} = 105$ Hz

So the first harmonic (lowest resonant frequency) is 105 Hertz. (Note that we were able to calculate this without taking into account the length of the string)