= Solution
At the first two specified times, the displacement profiles are
$$
\boxed{y(x,0)=y_0\cos kx,\qquad y\left(x,\frac{\pi}{2\omega}\right)=y_0\sin kx.}
$$
The second curve is the first shifted right by $\pi/(2k)=\lambda/4$. More generally a point of fixed wave <phase> satisfies $\omega t-kx=\text{constant}$, so it moves along $x=(\omega/k)t+\text{constant}$. Hence the wave travels in the positive $x$ direction with
$$
\boxed{c=\frac\omega k.}
$$
This <phase velocity> is distinct from the local transverse <velocity> $\partial y/\partial t$ of a string element.
For the <force> derivation take a short segment between $x$ and $x+\Delta x$. Let the tangent angles to the positive $x$ axis be $\theta_L$ and $\theta_R$. The left <tension> pulls along the negative tangent and the right <tension> along the positive tangent. With small slopes and approximately constant <tension> magnitude $T$, the horizontal components cancel to leading order and
$$
F_y=T\sin\theta_R-T\sin\theta_L\simeq T[y_x(x+\Delta x)-y_x(x)]\simeq T\Delta x\,y_{xx}.
$$
The segment <mass> is $\rho\Delta x$, where $\rho$ is \b[<mass> per unit length], not volume density. <Newton's second law> therefore gives
$$
\boxed{\rho y_{tt}=Ty_{xx},\qquad y_{tt}=c^2y_{xx},\qquad c=\sqrt{\frac T\rho}.}
$$
The approximation neglects bending stiffness, damping and changes in <tension> caused by finite-<amplitude> stretching.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017-string-waves.png]
{title=String-element forces, a right-moving sinusoid and two Gaussian-pulse initial conditions}
Converting the supplied density gives $\rho=0.020\,\mathrm{kg\,m^{-1}}$, so
$$
c=\sqrt{\frac{1000}{0.020}}=223.6\,\mathrm{m\,s^{-1}},\qquad \boxed{f=\frac c\lambda=\frac{223.6}{0.5}=447\,\mathrm{Hz}.}
$$
For the pulse, define $g(x)=Ae^{-ax^2/2}$ with $a>0$. The given initial <velocity> is $-cg'(x)$, exactly the <velocity> associated with a right-moving displacement $g(x-ct)$. Therefore
$$
\boxed{y(x,t)=A\exp\left[-\frac a2(x-ct)^2\right].}
$$
It moves right without changing shape in this nondispersive model. With the same initial displacement but zero initial transverse <velocity>, <D'Alembert's formula> instead gives the <Gaussian pulse splitting from zero initial velocity>
$$
\boxed{y(x,t)=\frac A2\left\{e^{-a(x-ct)^2/2}+e^{-a(x+ct)^2/2}\right\}.}
$$
The two half-<amplitude> pulses travel in opposite directions. Their initial velocities cancel, while their initial displacements add to the prescribed <Gaussian>. The two initial-value problems have different kinetic energies; splitting into half-height pulses does not contradict <energy> conservation. These forms assume a string long enough that reflections from its ends have not returned.
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