Answer:
Force is 161.27 N and frequency of vibration when the string vibrates in three segments is 660 Hz .
Step-by-step explanation:
Given :
Frequency , f = 220 Hz .
Length of wire , L = 70 cm = 0.7 m .
Mass of wire ,
![m = 1.20\ g=(1.2)/(1000)=1.2* 10^(-3)\ kg .](https://img.qammunity.org/2021/formulas/physics/college/e5j3oo43ijuz5gzdilq8w3vy92qve1z1wq.png)
a ) We know, frequency in string is :
![f=(1)/(2L)\sqrt{(F)/(\mu)}](https://img.qammunity.org/2021/formulas/physics/college/y37jp02h0fnm4dslgu35w5r0npc00jdlt9.png)
Therefore ,
.... equation 1.
Here ,
is mass per unit length .
So,
![\mu=(m)/(L)=(1.2* 10^(-3)\ kg)/(0.7\ m)=1.7* 10^(-3)\ kg/m.](https://img.qammunity.org/2021/formulas/physics/college/r8h7a3oqdk0ums88jwgc3vuu4covyd56re.png)
Putting value of
in equation 1.
We get ,
![F=4* 1.7* 10^(-3) * 0.7^2 * 220^2=161.27\ N.](https://img.qammunity.org/2021/formulas/physics/college/vdgi9bup44u54uz2c4negorw51aueog3bc.png)
b) We know , frequency of when n segment are in string :
![f_n=nf_1](https://img.qammunity.org/2021/formulas/physics/college/7kp5m1hbdzfsimdqh0kbdxaghuw3znm1wh.png)
For , n = 3
![f_3=3f_1\\\\f_3=3* 220\\\\f_3=660\ Hz.](https://img.qammunity.org/2021/formulas/physics/college/o53hiim9mk38hx90pbho3oyfa2vjon4r5e.png)
Hence , this is the required solution.