Question

A simple instrument consists of two metal strings stretched parallel to one another and attached at both ends to boards perpendicular to the strings. The tension is the same in both strings, and the length of the strings is ` = 35.0 cm. The first string has a mass m = 8.00 g and generates middle C (frequency f = 262 Hz) when vibrating in its fundamental mode.

a) What is the tension in the first string?

b) If the second string generates A-sharp (frequency 466 Hz) when vibrating in its fundamental mode, what is its mass?

c) A fret is placed just under the two strings, but not touching them. The fret is perpendicular to the strings, parallel to the boards. When the first string is pressed against the fret and plucked, the note emitted (in the fundamental mode of oscillation) is C-sharp (277 Hz). What is the distance between the fret and the closest end of the instrument?

d) When the second string is pressed against the fret and vibrates in its fundamental mode, what frequency is produced?

Answer 1

frequency f = v / $\lambda$ ; and velocity of sound is
T W =
where T is tension and $\mu$ = m / L is linear density of string and for fundamental mode L = $\lambda$ / 2

The fundamental frequency of sound in a stretched string is given by $f_{0}=\frac{1}{2L}\sqrt{\frac{TL}{m}}$ ;

4mLf6 = 1

The Tension in the first string is T = 4 x 0.008 x 0.35 x 2622 = 768.8 N

b) Since tension in the two strings is same but masses and frequencies are different then $4m_{1}Lf_{1}^{2}=T=4m_{2}Lf_{2}^{2}$ => $m_{2}=m_{1}\frac{f_{1}^{2}}{f_{2}^{2}}$ => $m_{2}=8 g\left (\frac{262}{466} \right )^{2}$

The mass of the second string is m2 = 2.53 g = 0.00253 kg

c) f1 L1 = f2 L2 => 262 x 0.35 = 277 x  L2​​​​​​​ => L2​​​​​​​ = 0.331 m

The distance between the fret and the closest end of the instrument is 0.35 - 0.331 = 0.019 m

d) For second string the fundamental frequency when string is pressed at fret is   f1 L1 = f2 L2

466 x 0.35 =  f2 x 0.331 => f2 = 492.75 Hz