Waves on a string, ripples on a pond are transverse waves generated by mechanical energy and in the simplest form oscillations. Sound also is a form of energy and basically oscillations. What makes sound unique in the fact that it is longitudinal ?
[Physics] Why are sound waves longitudinal even though they are mechanical energy
acousticsenergygaswaves
Related Solutions
Imagine a metal rod and you hit one end of it with a hammer.
A compression pulse travels down the rod and back again after reflection at the other end and the cycle of reflections is repeated - this is a longitudinal wave motion.
However as the pulse travels down the rod I would imagine that the walls of rod bulge out and then return, so this is equivalent to a transverse wave.
That bulging out then compresses and rarefies the air in the vicinity of the rod and so you get longitudinal waves in air produced.
A fake derivation
We can rather easily compute a horizontal velocity for the string fi we assume that the total velocity vector is everywhere normal to the string (this assumption is not always valid, see below). The following picture then illustrates the computation:
Take two infinitesimally separated points $x$ and $x+\mathrm{d}x$ and let the wave motion be $\varphi(x,t)$. The vertical/transverse velocity is $v_\text{vert} = \partial_t \varphi(x,t)$, and the horizontal component is $v_\text{hor} = -v_\text{vert}\tan(\vartheta)$, where $\vartheta$ is the angle between the normal and the vertical, and the minus sign is because if we measure $\vartheta$ in the usual counterclockwise direction then the horizonal velocity points to $-x$ for small $\vartheta$. Now $\tan(\vartheta)$ is $\frac{\varphi(x+\mathrm{d}x) - \varphi(x)}{\mathrm{d}x} = \partial_x\varphi(x)$ , so we get $$ v_\text{hor} = -\partial_t\varphi\partial_x\varphi$$ and if you plug in the sinusoidal solution and take the time average you get exactly the same result as for longitudinal waves. However, you might protext - the transverse wave equation was derived assuming no longitudinal motion, and this computation just blatantly assumes something different.
A Lagrangian derivation
Oddly enough, the result of the above computation is the correct momentum for a pure transverse wave. The Lagrangian of a transverse wave is $$ L = \frac{1}{2}\rho (\partial_t\varphi)^2 - \frac{1}{2}\tau(\partial_x\varphi)^2$$ and translation invariance gives us a momentum density $$ T_{xt} = \partial_x L \partial_t \varphi = - \rho\partial_x\varphi\partial_t\varphi$$ which is conserved by Noether's theorem.
The actual answer
In reality, there are no purely transverse waves on a string, there will always be secondary longitudinal waves generated when trying to excite it purely transversely. The "true" momentum of a realistic "transverse" wave is rather half of the theoretical prediction, i.e. $\frac{1}{2}\rho\partial_t\varphi\partial_x\varphi$, for more on this see "The missing wave momentum mystery"[pdf link] by Rowland and Pask.
Best Answer
In order for mechanical waves to propagate there needs to be some form of "restoring force" that tries to bring the system back to equilibrium. For longitudinal waves in gases this restoring force is supplied by pressure in the medium. However, there is no restoring force in a gas for bulk shear movement of the gas particles. Therefore, there is no transverse wave propagation in gases.
However, as mentioned in the comments, sound can also propagate transversely through solids, where there is a restoring force due to shear movement. The fact that both bulk transverse and longitudinal waves can propagate through solids, but only bulk longitudinal waves can propagate through fluids, is important in studying seismic waves, since the Earth is made up of both solid and fluid components.