Answer:
0.0549 m
Explanation:
Given that
equation y(x,t)=Acos(kx−ωt)
speed v = 8.5 m/s
amplitude A = 5.5*10^−2 m
wavelength λ = 0.5 m
transverse displacement = ?
v = angular frequency / wave number
and
wave number = 2π/ λ
wave number = 2 * 3.142 / 0.5
wave number = 12.568
angular frequency = v k
angular frequency = 8.5 * 12.568
angular frequency = 106.828 rad/sec ~= 107 rad/sec
so
equation y(x,t)=Acos(kx−ωt)
y(x,t)= 5.5*10^−2 cos(12.568 x−107t)
when x =0 and and t = 0
maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))
maximum y(x,t)= 5.5*10^−2 m
and when x = x = 1.52 m and t = 0.150 s
y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )
y(x,t)= 5.5*10^−2 × (0.9986)
y(x,t) = 0.0549 m
so the transverse displacement is 0.0549 m
