0
$\begingroup$

The resonance apparatus (shown below) is used to calculate the speed of sound using constructive and destructive interference of sound waves.

While conducting the experiment, we continuously vary the length of water column until we achieve resonance.

We alter the length of water in the water column in the following way-

  • We raise the water reservoir higher than it's original height. The water level in both the tubes (column and reservoir) is seen to increase.
  • We lower the water reservoir below it's original height. The water level in both the tubes (column and reservoir) is seen to decrease.

This doesn't seem intuitive at all.
- If the water increases in both tubes then where does this extra water come from?
- If the water decreases in both tubes then where does some of the water go?
- Shouldn't liquid stay at the same level? so that the same pressure is maintained?

Further Clarification:
If the length of the height in the water column (BC) be $h$. Then I expect to see the height of the water reservoir also $h$ so that they are at the same level.
Now as we raise the water reservoir, the water level should remain exactly the same (equal to $h$) level. But what happens is the height of the water column and the level of water in the reservoir both increase and become more than $h$.
The opposite happens when we decrease the height of the water reservoir.

All help is appreciated. Resonance Tube Apparatus

$\endgroup$
  • 1
    $\begingroup$ I think this is simply a question of reference frame. "increase" or "decrease" with respect to what? I think the reference here is "outside observer" not "center of the contraption". If you lift the reservoir the top of the water column goes up with respect to the outside world but less then the reservoir itself. So arguably it goes down with respect to the "center of the contraption" $\endgroup$ – Hilmar May 2 '18 at 17:27
1
$\begingroup$

If both sides are open to the atmosphere as in your diagram, the water stays at the same level on the two sides. The water level moves up or down depending on whether you move the reservoir up or down.

No water appears or disappears. The volume of water stays the same. What changes is the shape of the vessel. Instead of having left and right sides of equal length (as in the diagram on the left below), you are making the reservoir side (on the right) shorter and the cylinder side (left) longer, or vice versa. More importantly you are altering the 'depth' of the connected vessels : the bottom of the connecting tube is moving down or up.

enter image description here

For example, suppose the water level is 5cm below the tops of the tube on both sides and 15cm above the bottom of the connecting tube. If you lower the reservoir (right side) by 10cm, the water level is lowered by 5cm. Relative to the tops of the tube, the water rises on the right by 5cm (so it is now level with the top) and lowers on the left by 5cm (so it is now 10cm from the top).

This operation can be done in 2 steps :

  1. Cut off the top 5cm from the right side and add it to top of the left side, changing it from a U shape to a backward J shape. This alters the shape of the tube, and changes the water level relative to the tube but not relative to the laboratory.

  2. Lower the altered tube by 5cm. This changes the water level relative to the laboratory.

The shape of the U tube has altered. The shape of the water has remained the same. All that has happened is that it is now 5cm lower. The water level is still 15cm above the bottom of the tube. Neither the volume of the tube nor that of the water have changed.

$\endgroup$
  • $\begingroup$ +1 Thanks for the answer; I think your answer indeed makes sense! $\endgroup$ – SmarthBansal May 2 '18 at 16:51
  • $\begingroup$ (Clarifying) As we lower the reservoir, the main water column decreases. And the water in the reservoir though increases, appears to decrease as it was lowered from it's initial height. (right?) $\endgroup$ – SmarthBansal May 2 '18 at 16:59
  • 1
    $\begingroup$ That is correct. Lowering the reservoir, the water level on both sides goes down relative to its previous position. But relative to the top of the reservoir it goes up. $\endgroup$ – sammy gerbil May 2 '18 at 17:04
2
$\begingroup$

The water takes up a fixed volume $V$. It will occupy the tubes up to a height that allows that encloses that volume.

By manipulating the tubes, you are changing the cross-sectional area that is lowest. When the average cross-sectional area is lower, the water has to rise to a higher level. When the average cross-sectional area is higher, the water falls to a lower level.

This is all assuming static analysis (the water has time to reach equilibrium). If you're moving things around quickly, then the inertia of the water can change the observations.

$\endgroup$
  • $\begingroup$ +1 Thanks for the answer! Though I'm not sure how manipulating the tubes changes the cross-sectional area? $\endgroup$ – SmarthBansal May 2 '18 at 16:53
  • 1
    $\begingroup$ The reservoir has a wide area, the tube below it has a small area. If you lift the reservoir, you're lowering the average cross-sectional area in the lowest part of the tube (which is where the water occupies). $\endgroup$ – BowlOfRed May 2 '18 at 16:58
  • $\begingroup$ not sure I still follow; Is it like when we lift the reservoir the tube at the bottom gets a little bent and that constricts the area of crossection? $\endgroup$ – SmarthBansal May 2 '18 at 17:09
  • $\begingroup$ No. The tube itself retains its shape. But we are only interested in the lowest portion. So the average cross-section of the lowest portion depends on the orientation. Lifting a wide section higher means that the lowest portion now is thinner (not because the tube changed, but because what comprises the "lowest section" has changed). $\endgroup$ – BowlOfRed May 2 '18 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.