The vibrations of a string of length 60 cm fixed at both ends are represented by the equation `y=4sin((pix)/15) cos (96 pi t)`, where x and y are in cm and t in seconds.
(a)What is the maximum displacement of a point at `x = 5cm`?
(b)Where are the nodes located along the string?
(c)What is the velocity of the particle at x=7.5cm and t=0.25s?
(d)Write down the equations of the component waves whose superposition gives the above wave.

Options

A. The maximum displacement of a particle at `x=5cm` is `2sqrt(3) cm`.
B. The nodes located along the string are `15n` where ineger `n ` varies from 0 to 40.
C. The velocity of the particle at `x=7.5cm` at `t=0.25 sec` is zero
D. The equations of the component waves whose superpostion gives the above wave are `2 sin 2pi ((x)/(30)+48t),2 sin 2pi ((x)/(30)-48t)`

Answer

A, B, C, D

Explanation

`y=4sin (pi(x)/(15))cos 96pit`
At `x=5cm,y=4sin((pi)/(3))` and `y_(max)=2sqrt(3)cm`
Positions of nodes is given by equation
`sin ((pix)/(15)) = 0`
`rArr (pix)/(15)=npi`
At `x=7.5cm` and `t=0.25sec.`
Velocity of the particle `=(dely)/(delt)=-344pisin((pix)/(15))sin (96pit)=0`
`y=4sin ((pix)/(15))cos (96pit)=2sin((pix)/(15)+96pit)+2sin((pix)/(15)-96pit)`