1.

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

Answer» Correct Answer - A::B::C::D
(i) Here amplitude, `A = 4sin((pix)/(15))`
`x = 5m`
`A = 4sin((pix)/(15)) = 4 xx 0.866 = 3.46cm`
(ii) Nodes are the position where `A = 0`
:. `sin((pix)/(15)) = 0 = sinnpi` :. `x = 15 n`
where `n = 0,1,2` `x = 15cm, 30cm, 60cm,....`
(iii)`y = 4sin((pix)/(15))cos(96pit)`
`v = (dy)/(dt) = 4sin ((pix)/(15))[ - 96pi sin(96pit)]`
At `x = 7.5cm`, `t = 0.25cm`
`v = 4sin((pi xx 7.5)/(15))[ - 96pi sin(96pi xx 0.25)]`
`= 4sin((pi)/(2))[- 96pi sin(24pi )] = 0`
(iv) `y = 4sin ((pix)/(15))cos[96pit]`
`= 2[(2 sin((pix)/(15))cos (96pit)]`
`= 2[ sin(96pit + (pix)/(15))-sin (96pit -(pix)/(15))]`
gt`= 2 sin(96pit + (pix)/(15))-2sin (96pit -(pix)/(15))]`
`= y_(1) + y_(2)`
where `y_(1) = -2sin(96pit + (pix)/(15))`
and `y_(2) = -2sin(96pit -(pix)/(15))`


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