A string oscillates according to the equation below. y \' = (0.60cm) sin[( /3 cm

Question

A string oscillates according to the equation below. y' = (0.60cm) sin[(/3 cm-1)x]cos[(35s-1)t]

(d) What is the transverse speed of a particle of the string at theposition x = 1.5 cm when t = 9/8 s?

There are other parts to the question that I got right I just amreally confused with this section.

Please Help!
y' = (0.60cm) sin[(/3 cm-1)x]cos[(35s-1)t]

(d) What is the transverse speed of a particle of the string at theposition x = 1.5 cm when t = 9/8 s?

There are other parts to the question that I got right I just amreally confused with this section.

Please Help!

Explanation / Answer

The transverse speed (y-direction) = 0 This is a standing wave and hence the y-coordinates areconstant for a given x. sin A cos B = 1/2 [sin (A + B) + sin (A - B)] sin x / 3 cos 35 t = sin / 3 (x + 105 t)+ sin / 3 (x - 105 t) which is the form y = sin k (x + v t) + sin k (x - v t) thesum of two waves traveling in opposite directions (a standingwave)

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