Question

A transverse wave is given by the following equation:

y (x,t) = (2.45 cm) cos [(0.420 rad/cm)x + (5.20 rad/s)t]

a) What are the wave's amplitude, frequency, period, and wavelength?

b) What is the direction of wave travel, and what is the speed?

c) What is the displacement of a particle at x = 5.00 m, at t = 1.00 min?

Hint: pay attention to units!

Answer 1

) Sonn- yout) = A sin (KX-W!) - 0 General egh of wque moving along tue x-axis where A = amplitude , w= frequency: Given that yorot) = 2.45 cm cos [0:42 radleyin 2 +5.2 ralls it 7 (82 Compairing above ean to ① amplitude (A) = 2.45 cm Ang 5.2 radis 2 TT 5.2 radis T = 1.21 sec Ang 0.42 ralcm k= 21 а ते 14.952 cm Ang (b) The wave is travelling along Eve? x- axis velocity of wque U) = Coefficient of t coollicient of a - 5.2 к 0.42 W ² 12 381 cm Is Are

© at x= 5m = 500 cm t = 1 min = 60 fee. Cut) = your 2.45.cm cos [0.42 x 45.2 t] = 2:45 cos [0.42X500 + 5.2 x 60 = 9.45 Cos 522 = 2.33 cm y (228) Are