Answer:
f_n= 80 Hz
Explanation:
For first four nodes there are five nodes and four antinodes and it is called a frequency of four nodes also called as frequency of fourth over tune
resonance frequency is given by
[tex]f_n= \frac{n}{2l}\sqrt{\frac{\tau}{\mu} }[/tex]
Given problem
L=1 m
τ= 16 N
μ= 0.01 kg/m = linear density of the string
for 4 nodes
n= 4
therefore, [tex]f_n= \frac{4}{2\times1}\frac{16}{0.01}[/tex]
f_n= 80 Hrz
Answer:
[tex]F_4 = 80 Hz[/tex]
Explanation:
For given 4 modes there 4 nodes and 4 anti-nodes
A strings with four modes are given in figure
[tex]Fn = \frac{n}{2L} \sqrt{\frac{\tau}{\mu}}[/tex]
from the data given in question we have
L = 1 m
[tex]\tau = 16 N[/tex]
[tex]\mu = 0.01 kg/m[/tex]
so resonant frequency is given as
[tex]Fn = \frac{4}{2\times 1} \sqrt{\frac{16}{0.01}}[/tex]
Fn = 80 Hz
[tex]F_4 = 80 Hz[/tex]
