Water waves in a lake travel 3.4 m in 1.8 s. The period of oscillation...

a)
$v=\frac{d}{t}=\frac{3.4}{1.8}=1.9m/s$
b)
$\lambda =vT=(1.9)\ast (1.1)=2.1m$
Result:
a) 1.9m/s
b) 2.1m

Given:
Distance s =3.4m
Time: t=1.8s
Period: T=1.1s;
Required:
a) Wave speed v;
b) Wavelength of the water wave $\lambda$ ;
Water waves spread out at a constant speed given as the ratio of distance traveled and the time needed to travel that distance.
We have an equation that relates the wave speed, wavelength, and frequency. 1The period can be found as the reciprocal of the frequency:
$v=\frac{s}{t}$...(1)
$v=\lambda f$...(2)
$T=\frac{1}{f}$...(3)
a) The speed of the water waves can be found using the first equation. Plugging in the numbers we have:
$v=\frac{s}{t}$
$=\frac{3.4m}{1.8s}$
$=1.9\frac{m}{s}$
$v=1.9\frac{m}{s}$
b) Solving for $\lambda$ we divide the second equation by f and combine by f and combine it with the third equation:
$\lambda =\frac{v}{f}$
=vT
$=1.9\frac{m}{s}\ast 1.1s$
=2.1m
$\lambda =2.1m$
Result:
a) $v=1.9\frac{m}{s}$
b) $\lambda =2.1m$