How Do Transverse and Longitudinal Waves Differ in a Slinky Experiment?

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Transverse and longitudinal waves exhibit distinct behaviors in a slinky experiment, particularly regarding amplitude and tension. The amplitude of transverse waves does not significantly affect their speed, while increased tension in the slinky typically results in faster wave propagation. For longitudinal waves, moving your hand back and forth creates compressions and rarefactions in the coils, with one complete vibration of the hand producing one compression and one rarefaction. Understanding these concepts is crucial for grasping wave dynamics in various mediums. The discussion highlights the importance of tension and hand motion in wave behavior.
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waves and sound

I have some questions about transverse and longitudinal waves...
In transverse wave, how does the amplitude of the waves affect its speed? we did a lab with a slinky but it was hard to see if it went faster or slower...
Also, how does the tension of the spring change its speed? I think i noticed the pulse going faster when the slinky was tenser but... why does it do that?
And in longitudinal waves... when you move your hand back and forth quickly at a uniform frequency with the slinky, what happens to the motion of each coils?
Lastly, how many complete vibration of your hand do u think is need to produce one compression and one rarefaction? I just don't quite understand what it means by hand vibrations...;;
 
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jnimagine said:
complete vibration of your hand

This would likely be one back and forth of your hand, as you send one pulse (either transverse or longitudinal) down the slinky.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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