A transverse wave propagates along the positive direction of the axis x. Calculate??

A transverse wave propagates along the positive direction of the axis x being the maximum value of y = 6 cm, λ = 8π cm, v = 48 cm / s and the displacement at x = 0, t = 0 is 2 cm. Calculate: a) the wave number b) the angular frequency c) The phase count d) write the expression of this wave and that of another wave with the same frequency, wavelength and amplitude but such that when added to the previous one of a similar amplitude equal to 9.6cm

May 31, 2018

OK, quite involved, but we’ll have a go, and have a stab at (d)

Explanation:

Assuming the y = 6 cm refers to the amplitude of the wave, and that the wave is sinusoïdal I get:

(a) the wavenumber, $n = \frac{1}{\lambda} = \frac{1}{8} \pi = 0.0398 c {m}^{-} 1$

(b) the angular frequency, $\omega = 2 \pi f = \frac{2 \pi v}{\lambda}$
So $\omega = \frac{2 \times \cancel{\pi} \times 48}{8 \cancel{\pi}} = 12 \frac{r a d}{s}$

(c) the phase constant, or phase angle, $\phi = \omega t$ so we’d need to know the time at which the phase constant is required ...

(d) $y = 6 \sin \left(\omega t + 2\right)$
The final part is ambiguous to me, is the total amplitude when summing the waves equal to 9.6cm? If so it will be of the form $y = 9.6 \sin \left(\omega t + 2\right)$ I think (all we are changing is the amplitude)

Apologies if I have sent you down the wrong track here - I’ll request a check before I post this answer.