Answer:
Force is 161.27 N and frequency of vibration when the string vibrates in three segments is 660 Hz .
Explanation:
Given :
Frequency , f = 220 Hz .
Length of wire , L = 70 cm = 0.7 m .
Mass of wire , [tex]m = 1.20\ g=\dfrac{1.2}{1000}=1.2\times 10^{-3}\ kg .[/tex]
a ) We know, frequency in string is :
[tex]f=\dfrac{1}{2L}\sqrt{\dfrac{F}{\mu}}[/tex]
Therefore ,
[tex]F=4\mu L^2f^2[/tex] .... equation 1.
Here , [tex]\mu[/tex] is mass per unit length .
So, [tex]\mu=\dfrac{m}{L}=\dfrac{1.2\times 10^{-3}\ kg}{0.7\ m}=1.7\times 10^{-3}\ kg/m.[/tex]
Putting value of [tex]\mu[/tex] in equation 1.
We get , [tex]F=4\times 1.7\times 10^{-3} \times 0.7^2 \times 220^2=161.27\ N.[/tex]
b) We know , frequency of when n segment are in string :
[tex]f_n=nf_1[/tex]
For , n = 3
[tex]f_3=3f_1\\\\f_3=3\times 220\\\\f_3=660\ Hz.[/tex]
Hence , this is the required solution.
