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The vibrations of a string of length 60c...

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.

Text Solution

Verified by Experts

The correct Answer is:
(i) `2 sqrt(3)cm`
(ii) `1.49 xx10^(-3)m`
(iii) 0
(iv) `y_(1)=2sin((pix)/(15)-96pit)and y_(2)=2sin((pix)/(15)+96pit)`

(i) `(y_(max))_(x=5cm)=4 sin((pix)/(15))=4sin((5pi)/(15))=2sqrt(3)`
(ii) `(2pi)/(lamda)=(pi)/(15)rArr lamda=30cm`
`:.` Position of nodes
`=15cm,30cm`….
(iii) `v=-4(96pi)sin((pix)/(15))sin(96pit)`
at`x=7.5 cm,t=0.25sec`
`v=-4 xx 96 pi sin((7.5pi)/(15)) sin((96pi)/(4))=0`
(iv) `Y=4 sin((pix)/(15))cos(96pit)`
`=2sin((pix)/(15)-96t)+2sin((pix)/(15)+96pit)`.
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