A. guitar string vibrates at 400 Hz and has a length of 0.8 m. Find the new frequency ifa) the string is lengthened to 0.9 m.b) the string is replaced with another string, the density of which is greater by a factor of two.c) the tension in the string is decreased by a factor of 0.4.d) the string is replaced with another string, the diameter of which is twice as large.

[tex]\begin{gathered} \frac{n_1}{n_2}=\frac{L_2}{L_1} \\ n_2=\frac{L_1\cdot n_1}{L_2} \\ n_2=\frac{0.8\cdot400}{0.9} \\ n_2=355.56\text{ Hz} \end{gathered}[/tex]

(b)

ρ is the density

[tex]\begin{gathered} n\propto\frac{1}{\sqrt{\rho}} \\ \rho_2=2\rho_1 \\ \frac{n_1}{n_2}=\sqrt{\frac{\rho_2}{\rho_1}} \\ \frac{n_1}{n_2}=\sqrt[]{\frac{2\rho_1}{\rho_1}} \\ \frac{n_1}{n_2}=\sqrt[]{2} \\ n_2=\frac{n_1}{\sqrt[]{2}}=\frac{400}{\sqrt[]{2}} \\ n_2=282.84\text{ Hz} \end{gathered}[/tex]

(c)

T is the tension

[tex]\begin{gathered} n\propto\sqrt[]{T} \\ T_2=T_1-0.4\cdot T_1 \\ T_2=0.6\cdot T_1 \\ \frac{n_2}{n_1}=\sqrt[]{\frac{T_2}{T_1_{}}} \\ \frac{n_2}{n_1}=\sqrt[]{\frac{0.6\cdot T_1}{T_1}} \\ \frac{n_2}{n_1}=\sqrt[]{0.6} \\ n_2=400\cdot\sqrt[]{0.6} \\ n_2=309.84\text{ Hz} \end{gathered}[/tex]

(d)

d is the diameter

[tex]\begin{gathered} n\propto d \\ d_2=2\cdot d_1 \\ \frac{n_1}{n_2}=\frac{d_2}{d_1} \\ \frac{n_1}{n_2}=\frac{2\cdot d_1}{d_1} \\ \frac{n_1}{n_2}=2 \\ n_2=\frac{n_1}{2}=\frac{400}{2} \\ n_2=200\text{ Hz} \end{gathered}[/tex]