Equation of stationary wave along a stretched string is given by y = 5sin frac{{π x}}{3}cos 40π t

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14.Waves and Sound

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The equation of stationary wave along a stretched string is given by $y = 5\sin \frac{{\pi x}}{3}\cos 40\pi t$ where $x$ and $y$ are in centimetre and $t$ in second. The separation between two adjacent nodes is .... $cm$

A

$6$

B

$4$

C

$3$

D

$1.5$

Solution

(c) Given $y = 5\sin \frac{{\pi x}}{3}\cos 40\,\pi t$

Comparing with $y = 2a\cos \frac{{2\pi vt}}{\lambda }\sin \frac{{2\pi x}}{\lambda }$$ \Rightarrow \lambda = 6\,cm.$

$\therefore$ The separation between adjacent nodes $ = \frac{\lambda }{2} = 3\,cm.$

Standard 11

Physics

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