had a small question, how to take out transverse speed of a wave @Irishboy123

Question

arindameducationusc · Answer

See attachment, only part b needed

arindameducationusc · Answer

Check point 2

arindameducationusc · Answer

@Michele_Laino

irishboy123 · Answer

the transverse speed is the speed perpendicular to wave travel and technically is the partial derivative wrt t

so for $y(x,y) = 2 \sin (4x - 2t)$

$v _{\perp} = \dfrac{\partial  }{\partial t}(2 \sin (4x - 2t))$

you can get the same result just by choosing a fixed point on the x axis, say , x = 0 and looking at $\dfrac{d  }{d t}(2 \sin (- 2t)$ instead.  as if it were a standing wave.  at all points you will get the same oscillation in the transverse direction so this works nicely

so that leaves us with $v_{\perp}= -4 \cos(-2t) = -4 \cos(2t)$,  and we have a wave whose velocity transverse alternates between 4 and -4 as cosine  alternates between -1 and 1.  

so max speed for that = 4

repeat for others....

arindameducationusc · Answer

Got it @IrishBoy123 
Thank you