## arindameducationusc one year ago had a small question, how to take out transverse speed of a wave @Irishboy123

1. arindameducationusc

See attachment, only part b needed

2. arindameducationusc

Check point 2

3. arindameducationusc

@Michele_Laino

4. IrishBoy123

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....

5. arindameducationusc

Got it @IrishBoy123 Thank you