Answer:
[tex]\frac{v_s}{v_l}=\frac{5}{6}=0.8333...[/tex]
[tex]\frac{T_s}{T_l}=\frac{25}{36}=0.69444...[/tex]
Explanation:
The frequency of the nth harmonic on a string is given by the formula:
[tex]f_n=\frac{nv}{2L}[/tex]
which can be written as:
[tex]v=\frac{2Lf_n}{n}[/tex]
Since the fundamental frequency means n=1, the ratio of the speed of waves on the shorter string ([tex]v_s[/tex]) to the speed of waves on the longer string ([tex]v_l[/tex]) will be:
[tex]\frac{v_s}{v_l}=\frac{2L_sf_{1s}}{2L_lf_{1l}}=\frac{L_s}{L_l}[/tex]
Since the frequencies are the same, and for our values is:
[tex]\frac{v_s}{v_l}=\frac{L_s}{L_l}=\frac{70cm}{84cm}=\frac{5}{6}=0.8333...[/tex]
The speed of the waves relates to the tension and mass per unit length by the formula:
[tex]v=\sqrt{\frac{T}{\lambda}}[/tex]
which can be written as:
[tex]T=v^2 \lambda[/tex]
The ratio of the tension in the shorter string to the tension in the longer string will be:
[tex]\frac{T_s}{T_l}=\frac{v_s^2 \lambda_s}{v_l^2 \lambda_l}=\frac{v_s^2}{v_l^2}=(\frac{v_s}{v_l})^2[/tex]
Since the mass per unit length are the same, and for our values is:
[tex]\frac{T_s}{T_l}=(\frac{v_s}{v_l})^2=(\frac{5}{6})^2=\frac{25}{36}=0.69444...[/tex]
