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Solution :

Let us compare the given equation with `y=A sin(omegat-kx+phi)`. <br> we find that `k=3pi rad//m` and `omega=10pi rad//s` <br> (a) The speed and direction of the wave are both specified by the vector wave velocity: <br> `vec(v)=f lambda hat(i)=(omega)/(k)hat(i)=(10pi rad//s)/(3pi rad//m)hat(i)=3.33hat(i) m//s` <br> (b) substituting `t=0` and `x=0.100 m`, we have `y=(0.350 m) sin (-0.300pi+0.250pi)` <br> `=(0.350 m) sin (-0.157)` <br> `=(0.350 m)(-0.156)=-0.0548 m=-5.48 cm` <br> note that when you take the sine of a quantity with no units, the quantity is not in degrees, but in radians. <br> (c ) The wavelength is <br> `lambda=(2pi rad)/(k)=(2pi rad)/(3pi rad//m)=0.667 m` <br> and the frequency is <br> `f(omega)/(2pi rad) (10pi rad//s)/(2pi rad)=5.00 Hz` <br> (d) The particle speed is `v_(y)=dely//delt= (0.350 m)(10pi rad)cos(10pi t-3pix+pi//4)` <br> The maximum occurs when the cosine term is `1:` <br> `v_(y.max) =(10pi rad//s)(0.350 m//s)=11.0 m//s`