Question

Consider a string of length 1m, fixed at both ends, with tension τ = 16 N and mass/length μ = 0.01 kg/m. For each of the first 4 modes, sketch the mode shape, label the nodes and antinodes and calculate the resonant frequency.

Answer 1

`F_4 = 80 Hz`

Step-by-step explanation:

For given 4 modes there 4 nodes and 4 anti-nodes

A strings with four modes are given in figure

`Fn = (n)/(2L) \sqrt{(\tau)/(\mu)}`

from the data given in question we have

L = 1 m

`\tau = 16 N`

`\mu = 0.01 kg/m`

so resonant frequency is given as

`Fn = (4)/(2* 1) \sqrt{(16)/(0.01)}`

Fn = 80 Hz

`F_4 = 80 Hz`

Answer 2

f_n= 80 Hz

Step-by-step 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

`f_n= (n)/(2l)\sqrt{(\tau)/(\mu) }`

Given problem

L=1 m

τ= 16 N

μ= 0.01 kg/m = linear density of the string

for 4 nodes

n= 4

therefore,
`f_n= (4)/(2*1)(16)/(0.01)`

f_n= 80 Hrz