A string of linear mass density `5.0 xx 10^-3` kg//m is stretched under a tension of 65 N between two rigid supports 60 cm apart.
(a) If the string is vibrating in its second overtone so that the amplitude at one of its antinodes is 0.25 cm, what are the maximum transverse speed and acceleration of the string at antinodes?
(b) What are these quantities at a distance 5.0 cm from an node?

Answer

A, B, C, D

Explanation

(a) In second overtone,

`l=(3lambda)/2`
or `lambda= (2l)/3 = (2 xx 60)/3 `
`= 40 cm = 0.4 m `
` k= (2pi)/lambda = (2pi)/0.4 = 5pi m^(-1)`
`v=sqrt(T/mu)=sqrt(65/(5.0xx10^-3))`
`=114 m//s`
`omega=kv=570 pi rad//s`
Maximum transverse speed at antinode`=A_0omega`
Here, `A_0`=amplitude of antinode=`0.25 cm =2.5xx10^-3 m`
`:.` Maximum speed=`(2.5xx10^-3)(570 pi) m//s`
`=4.48 m//s`
Maximum acceleration =`w^2A_0`
=`(570pi)^2(2.5xx10^-3) m//s^2`
`=8.0xx10^3 m//s^2`
(b) At a distance x from the node, the amplitude can be written as
`A=A_0 sin kx=(2.5xx10^-3)sin(5 pix)`metre
Here, x is in metres.
Therefore, at =`x=5.0 cm =5.0xx10^-2 m`
`A=(2.5xx10^-3) sin (5 pixx5.0xx10^2)`
`=1.8xx10^-3 m`
Maximum speed=`A omega`
`=(1.8 xx (10^-3)) (570pi) m//s`
=3.22 `m//s`
and maximum acceleration `= omega^2A`
`= (570pi)^2 (1.8 xx (10^-3))m//s^2`
`=5.8 xx (10^3) m//s^2` .