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The equation of the vibration of a wire ...

The equation of the vibration of a wire is ` y = 5 "cos"(pi x)/(3) sin 40 pi t ` , where x and y are given in cm and t is given in s. calculate the
distance between two closest points of the wire that are always at rest,

Text Solution

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` y = 5 "cos"(pir)/(3) sin 40 pi t = (5)/(2) * 2 sin 40 pi t cos (pir)/(3)`
` = (5)/(2) [ sin ( 40 pi t + (pir)/(3))+ sin ( 40 pi t - (pir)/(3))]`
`= (5)/(2) sin 40 pi (t + (x)/(120))+(5)/(2)sin 40 pi (t - (x)/(120))`
` y _(1) + y_(2)`
So, the resultant vibration is produced due to the superposition of two waves ` y _(1) and y_(2)` . comparing these two waves with the general equation, y ` = A sin omega (t - (x)/(V)) ` .
Wavelength, `lambda = (V)/ (n) = (V)/(omega // 2 pi) = (2 pi r)/(omega)`
The closest points that are always at test denote two consecutive nodes. So , the distance between them
`= (lambda)/(2) = (pi V)/(omega) = (pi xx 120)/(40 pi) = 3 cm ` .
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