The vibrations of a string of length `60cm` fixed at both ends are represented by the equation----------------------------
`y = 4 sin ((pix)/(15)) cos (96 pit)`
Where `x` and `y` are in `cm` and `t` in seconds.
(i) What is the maximum displacement of a point at `x = 5cm`?
(ii) Where are the nodes located along the string?
(iii) What is the velocity of the particle at `x = 7.5 cm` at `t = 0.25 sec`.?
(iv) Write down the equations of the component waves whose superpositions gives the above wave.

Explanation

Compare the given eqn. With standard form of eqn. of stationary wave:
`y=2rsin((2pix)/(lambda))cos((2pi)/(T)t)`
we find `2r=4, r=2cm, (2pi)/(lambda)=(pi)/(15), lambda=30cm.`
Hence the string will vibrate in 4 segments, with each segment `=(lambda)/(2)=(30)/(2)=15cm` as shown in figure,
The nodes are at 0,15cm, 30cm, 45cm and 60cm as shown in figure
(ii) Velocity of particle
`upsilon=(dy)/(dt)=-4sin((pix)/(15))sin96pitxx96pi`
At `x=7.5 cm and t =0.25s`
`upsilon=-4xx96pisinpi//2sin(96pixx0.25)=Zero`
(iii) The component waves of the given stationary wave are
`y_(1)=2sin((pix)/(15)+96pit),` and
`y_(2)=2sin((pix)/(15)-96pit)`