Question

The portion of string between the bridge and upper end of the fingerboard (the part of the string that is free to vibrate) of a certain musical instrument is 60.0 cm long (0.6 m) and has a mass of 2.23 g (0.00223kg). The string sounds an A4 note (440 Hz) when played. Where must the player put a finger (at what distance from the bridge) to play a D5 note (587 Hz)? For both notes, the string vibrates in its fundamental mode. Without retuning, is it possible to play a G4 note (392 Hz) on this string?

Answer 1

Answer:

Part(a) : The required distance is 0.15 m.

Part(b) : Yes it is possible.

Explanation:

Part (a):

We know that, for a string to generate $n^{th}$ harmonic, it has to vibrate with a frequency ($f_{n}$) given by

$f_{n} = \dfrac{V}{\lambda_{n}} = \dfrac{nV}{2L}$

where '$V$' is the velocity of the velocity of the wave, '$\lambda_{n}$' is the wavelength of the wave for the $n^{th}$ harmonic, '$L$' is the length of the string.

In order to produce A4 note (440 Hz), i.e., $4^{th}$ harmonic on the string, if '$L_{4}$' be the required length of the string, then

$&&L_{4} = \frac{4 V}{2 f_{4}}\\&or,& V = \frac{L_{4}f_{4}}{2} = \frac{0.6 m \times 440 Hz}{2} = 132 m s^{-1}$

So, to generate a higher harmonic D5 (587 Hz), i.e., $5^{th}$ harmonic if we consider '$L_{5}$' m be distance from the bridge where the player has to put a finger, then

$&&V = \dfrac{(0.6 - L_{5}) \times f_{5}}{2}\\&or,& 0.6 - L_{5} = \dfrac{2 \times V}{f_{5}}\\&or,& L_{5} = 0.6 - \dfrac{2 \times V}{f_{5}} = 0.6 - \dfrac{2 \times 132}{587} = 0.15 m$

Part (b):

In order to produce G4 note (392 Hz) which is lower harmonic without retuning, the player has to make such an arrangement that will reduce the vibration of the string which in turn increase the waelength of the wave or decrease the frequency.