Question

A particular guitar wire is 30.0 cm long and vibrates at a frequency of 196 Hz in fundamental mode when no finger is placed on it. The next higher notes on the scale are 220 Hz, 247 Hz, 262 Hz and 294 Hz. How far from the end of the string must the finger be placed to play these notes ?

Answer 1

For 220 Hz: 3.27 cm For 247 Hz: 6.19 cm For 262 Hz: 7.56 cm For 294 Hz: 10.0 cm

Answer 2

The fundamental frequency ($f_1$) of a string of length $L$ fixed at both ends is given by the formula:

$f_1 = \frac{v}{2L}$

where $v$ is the speed of the wave on the string. The wave speed $v$ is determined by the tension ($T$) and linear mass density ($\mu$) of the string, $v = \sqrt{\frac{T}{\mu}}$. Since the same guitar wire is used and the tension is assumed to be constant, the wave speed $v$ is constant.

Let $L_0$ be the original length of the guitar wire and $f_0$ be its fundamental frequency.

$L_0 = 30.0$ cm $f_0 = 196$ Hz

So, the wave speed is $v = 2L_0 f_0$.

When a finger is placed on the string at a distance $x$ from one end, the vibrating length of the string becomes $L' = L_0 - x$. The finger acts as a new fixed point, and the string segment of length $L'$ vibrates in its fundamental mode to produce the desired note. The frequency of this fundamental mode is $f'$.

$f' = \frac{v}{2L'}$

Substitute $v = 2L_0 f_0$:

$f' = \frac{2L_0 f_0}{2(L_0 - x)} = \frac{L_0 f_0}{L_0 - x}$

We need to find the distance $x$ for each given frequency $f'$. Rearranging the formula to solve for $x$:

$L_0 - x = \frac{L_0 f_0}{f'}$ $x = L_0 - \frac{L_0 f_0}{f'} = L_0 \left(1 - \frac{f_0}{f'}\right)$

Now, we calculate $x$ for each of the given frequencies:

1. For $f'_1 = 220$ Hz:

$x_1 = 30.0 \text{ cm} \left(1 - \frac{196 \text{ Hz}}{220 \text{ Hz}}\right) = 30.0 \left(1 - \frac{49}{55}\right) = 30.0 \left(\frac{55 - 49}{55}\right) = 30.0 \left(\frac{6}{55}\right) = \frac{180}{55} = \frac{36}{11} \approx 3.27 \text{ cm}$.

2. For $f'_2 = 247$ Hz:

$x_2 = 30.0 \text{ cm} \left(1 - \frac{196 \text{ Hz}}{247 \text{ Hz}}\right)$. Note that $247 = 13 \times 19$.

$x_2 = 30.0 \left(\frac{247 - 196}{247}\right) = 30.0 \left(\frac{51}{247}\right) = \frac{1530}{247} \approx 6.19 \text{ cm}$.

3. For $f'_3 = 262$ Hz:

$x_3 = 30.0 \text{ cm} \left(1 - \frac{196 \text{ Hz}}{262 \text{ Hz}}\right) = 30.0 \left(1 - \frac{98}{131}\right)$. Note that 131 is prime.

$x_3 = 30.0 \left(\frac{131 - 98}{131}\right) = 30.0 \left(\frac{33}{131}\right) = \frac{990}{131} \approx 7.56 \text{ cm}$.

4. For $f'_4 = 294$ Hz:

$x_4 = 30.0 \text{ cm} \left(1 - \frac{196 \text{ Hz}}{294 \text{ Hz}}\right) = 30.0 \left(1 - \frac{2 \times 98}{3 \times 98}\right) = 30.0 \left(1 - \frac{2}{3}\right) = 30.0 \left(\frac{1}{3}\right) = 10.0 \text{ cm}$.

The distances from the end of the string where the finger must be placed to play the notes with frequencies 220 Hz, 247 Hz, 262 Hz, and 294 Hz are approximately 3.27 cm, 6.19 cm, 7.56 cm, and 10.0 cm, respectively.

Explanation:

The frequency of the fundamental mode of a vibrating string fixed at both ends is inversely proportional to its length ($f_1 \propto 1/L$). When a finger is placed on the string at a distance $x$ from one end, the vibrating length is reduced to $L' = L_0 - x$. The new frequency is the fundamental frequency corresponding to this shorter length. Using the relationship $f' = \frac{f_0 L_0}{L'}$, we can solve for the required distance $x = L_0 - L' = L_0 - \frac{f_0 L_0}{f'} = L_0(1 - \frac{f_0}{f'})$ for each target frequency.