= Solution
For opposite propagation directions on a uniform string, write the equal-frequency <transverse waves> as
$$
y_1=A\cos(kx-\omega t+\alpha),\qquad
y_2=A\cos(kx+\omega t+\beta).
$$
The cosine addition identity gives
$$
\boxed{y_1+y_2=2A\cos\left(kx+\frac{\alpha+\beta}{2}\right)
\cos\left(\omega t+\frac{\beta-\alpha}{2}\right).}
$$
This is a <standing wave>: the spatial pattern is fixed while its amplitude oscillates in time. Nodes occur at the zeros of the first cosine and are separated by $\pi/k$, half a <wavelength>; antinodes have displacement amplitude $2A$. The two equal opposite travelling waves carry equal opposite average <energy flux>, so the net average transported <power> is zero.
For the same propagation direction, the same identity instead gives
$$
\boxed{A\cos(kx-\omega t+\alpha)+A\cos(kx-\omega t+\beta)
=2A\cos\frac{\alpha-\beta}{2}\,
\cos\left(kx-\omega t+\frac{\alpha+\beta}{2}\right).}
$$
For a nonzero resultant this is another travelling wave, with fixed <amplitude> and moving constant-phase points, not a nontrivial <standing wave>. If the prefactor vanishes, the waves cancel identically; the zero disturbance is a trivial exception rather than a standing oscillation. The argument assumes propagating waves with $k,\omega>0$ on the same uniform string.
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