A transverse wave is represented by y = Asin( \omega t - kx). For what value of the wavelength is the wave velocity equal to the maximum particle velocity ?

Waves question

2010 · Q183

Waves question

2010 · Q183

View written solutionFree

Correct answer: C

y = A sin ( $\omega$ t–kx)

Particle velocity,

${v_p} = {{dy} \over {dt}} = A\omega \cos \left( {\omega t - kx} \right)$

$\therefore {v_{p\,\max }} = A\omega$

wave velocity = ${\omega \over k}$

$\therefore$ $A\omega = {\omega \over k}$

i.e, $A = {1 \over k}$ But $k = {{2\pi } \over \lambda }$

$\therefore \lambda = 2\pi A$

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Waves question

2010 · Q183

Waves question

2010 · Q183

View written solutionFree

Correct answer: C

y = A sin ( $\omega$ t–kx)

Particle velocity,

${v_p} = {{dy} \over {dt}} = A\omega \cos \left( {\omega t - kx} \right)$

$\therefore {v_{p\,\max }} = A\omega$

wave velocity = ${\omega \over k}$

$\therefore$ $A\omega = {\omega \over k}$

i.e, $A = {1 \over k}$ But $k = {{2\pi } \over \lambda }$

$\therefore \lambda = 2\pi A$