Question

The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 400 g. The vibrating section of the string is 1.90 m long. What tension is needed to tune this string properly?

Short answer

The tension needed to tune the given string is$2184N$.

Step-by-step solution

Given information

The frequency of the lowest note on a piano is $f_1=27.5Hz$.

The mass of the string isrole="math" localid="1650029973440" $mass=400g=0.400kg$.

Wavelength and wave speed

As the vibrating section of the string is $1.9m$long, the two ends of this vibrating wire are fixed, and the string is vibrating in the fundamental harmonic with wavelength

$$\begin{gathered} {\lambda }_1=2L \\ =2\times 1.9 \\ =3.8m \end{gathered}$$

The wave speed along the string is

$$\begin{gathered} v=f_1{\lambda }_1 \\ =27.5\times 3.8 \\ =104.5m/s \end{gathered}$$

Tension in the string

The linear density is given by

$$\begin{gathered} \mu =\frac{mass}{length} \\ =\frac{0.400}{2.00} \\ =0.2kg/m \end{gathered}$$

The wave speed is also given by role="math" localid="1650029766897" $v=\sqrt{\frac{T}{\mu }}$.

Substitute the given values

$$\begin{gathered} 104.5=\sqrt{\frac{T}{0.2}} \\ T=104.5^2\times 0.2 \\ =2184N \end{gathered}$$

Therefore, the tension in the string is$2184N$.