One end of a long rope is tied to a fixed vertical pole. The rope is stretched horizontally with a tension `8 N` . Let us consider the length of the rope to be along x-axis. A simple harmonic oscillator at `x=0` generates a transverse wave of frequency `100 Hz` and amplitude `2 cm` along the rope. Mass of a unit length of the rope is `20 g//m`. ignoring the effect of gravity, answer the following questions.
maximum magnitude of transverse acceleration of any point on the rope will be nearly

Options

A. `7888 m//s^(2)`
B. `8244 m//s^(2)`
C. `9277 m//s^(2)`
D. `3333 m//s^(2)`

Answer

A

Explanation

`v=sqrt((T)/(mu))=20 m//s`
`lambda=(v)/(f)=(20)/(100)=0.2 m=20 cm`
and `k=(2pi)/(lambda)=10pi, omega=2pif=200pi`
So `y=-0.2cos(10pix-200pit)`
wave velocity is constant for a medium but particle velocity keeps changing.
as `v=t'=4pi sin(10pix-200pit)`
`(d^(2)y)=-0.2x(200pi)^(2) cos(10pix-200pit)`
For `a_(max)=-aomega^(2)=-7888 m//s^(2)`
`|a_(max)|=7888 m//s^(2)`
As frequency doubles, `lambda` becomes half, speed of wave remains same.